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Exponential and Logarithmic Equations

Use like bases to solve exponential equations.

Exponential and Logarithmic Equations

Solve Exponential Equations. (IA 10.2.2)
Solve Logarithmic Equations. (IA 10.3.4)

Equations that include an exponential expression \({a}^{x}\) are called exponential equations. There are two types of exponential equations: those with the common base on each side, and those without a common base.

Type 1: Possible common base on each side: Use properties of exponents to rewrite each side with a common base. Use base-exponent property to set exponents equal to each other and solve for x.

Type 2: No possible common base: Use properties of exponents to rewrite each side in terms of one exponential expression. Take the log or ln of each side and use the power rule to bring down the power. Solve the remaining equation for x.

Example

Try it.

Solve: \({3}^{2x-5}=27.\)

Solution

Is here a common base?	Yes, both 3 and 27 can be rewritten as powers of 3.
Write both sides of the equation with the same base.	\({3}^{2x-5}={3}^{3}\)
Since the bases are the same, the exponents must be equal.
Write a new equation by setting the exponents equal.	\(2x-5=3\)
Solve the equation.	\(2x=8\)
	\(x=4\)
Check the solution by substituting x=4 into the original equation.	\({3}^{2(4)-5}=27\)
	\(27=27,\) True

Example

Try it.

Solve \(3{e}^{x+2}=24\) . Find the exact answer and then approximate it to three decimal places.

Solution

Rewriting with a common base is not possible.
Isolate the exponential by dividing both sides by 3.	\({e}^{x+2}=8\)
Take the natural logarithm of both sides.	\(\ln {e}^{x+2}=\ln 8\)
Use the Power Property to get the x as a factor, not an exponent.	\((x+2)\ln e=\ln 8\)
Use the property \(\ln e=1\) to simplify.	\(x+2=\ln 8\)
Solve the equation. Find the exact answer.	\(x=\ln 8-2\)
Approximate the answer.	\(x=0.079\)

Solve. Find the exact answer and then approximate it to three decimal places.

Try it.

\({4}^{2x-3}=\frac{1}{16}\)

Try it.

\(5({3}^{x})=20\)

Condensed — the full section is in OpenStax College Algebra 2e.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers \(b,\) \(S,\) and \(T,\) where \(b>0,\ b\ne 1,\) \({b}^{S}={b}^{T}\) if and only if \(S=T.\)

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation \({3}^{4x-7}=\frac{{3}^{2x}}{3}.\) To solve for \(x,\) we use the division property of exponents to rewrite the right side so that both sides have the common base, \(3.\) Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for \(x\):

\[\begin{array}{lll}{3}^{4x-7} & =\frac{{3}^{2x}}{3} & \\ {3}^{4x-7} & =\frac{{3}^{2x}}{{3}^{1}} & {\text{Rewrite 3 as 3}}^{1}. \\ {3}^{4x-7} & ={3}^{2x-1} & \text{Use the division property of exponents}\text{.} \\ 4x-7 & =2x-1\ \ & \text{Apply the one-to-one property of exponents}\text{.} \\ 2x & =6 & \text{Subtract 2}x\ \text{and add 7 to both sides}\text{.} \\ x & =3 & \text{Divide by 2}\text{.}\end{array}\]

Example

Try it.

Solve \({2}^{x-1}={2}^{2x-4}.\)

Solution

\[\begin{array}{lllll}{2}^{x-1}={2}^{2x-4} & \text{The common base is }2. \\ x-1=2x-4\begin{array}{llll} & & & \end{array} & \text{By the one-to-one property the exponents must be equal}. \\ x=3 & \text{Solve for }x.\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since \(\log (a)=\log (b)\) is equivalent to \(a=b,\) we may apply logarithms with the same base on both sides of an exponential equation.

Example

Try it.

Solve \({5}^{x+2}={4}^{x}.\)

Solution

\[\begin{array}{lllll}{5}^{x+2}={4}^{x} & \text{There is no easy way to get the powers to have the same base}. \\ \ln {5}^{x+2}=\ln {4}^{x} & \text{Take ln of both sides}. \\ (x+2)\ln 5=x\ln 4 & \text{Use laws of logs}. \\ x\ln 5+2\ln 5=x\ln 4 & \text{Use the distributive law}. \\ x\ln 5-x\ln 4=-2\ln 5 & \text{Get terms containing}\ x\ \text{on one side, terms without}\ x\ \text{on the other}. \\ x(\ln 5-\ln 4)=-2\ln 5 & \text{On the left hand side, factor out an }x. \\ x\ln (\frac{5}{4})=\ln (\frac{1}{25})\begin{array}{llll} & & & \end{array} & \text{Use the laws of logs.} \\ x=\frac{\ln (\frac{1}{25})}{\ln (\frac{5}{4})} & \text{Divide by the coefficient of}\ x.\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation \({\log }_{b}(x)=y\) is equivalent to the exponential equation \({b}^{y}=x.\) We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation \({\log }_{2}(2)+{\log }_{2}(3x-5)=3.\) To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x:\)

\[\begin{array}{lllll}{\log }_{2}(2)+{\log }_{2}(3x-5)=3 & \\ {\log }_{2}(2(3x-5))=3 & \text{Apply the product rule of logarithms.} \\ {\log }_{2}(6x-10)=3 & \text{Distribute}. \\ {2}^{3}=6x-10 & \text{Apply the definition of a logarithm}. \\ 8=6x-10\begin{array}{llll} & & & \end{array} & \text{Calculate }{2}^{3}. \\ 18=6x & \text{Add 10 to both sides}. \\ x=3 & \text{Divide by 6}.\end{array}\]

Example

Try it.

Solve \(2\ln x+3=7.\)

Solution

\[\begin{array}{ll}2\ln x+3=7 & \\ 2\ln x=4 & \text{Subtract 3}. \\ \ln x=2 & \text{Divide by 2}. \\ x={e}^{2} & \text{Rewrite in exponential form}.\end{array}\]

Example

Try it.

Solve \(2\ln (6x)=7.\)

Solution

\[\begin{array}{ll}2\ln (6x)=7 & \\ \ln (6x)=\frac{7}{2} & \text{Divide by 2}. \\ 6x={e}^{(\frac{7}{2})} & \text{Use the definition of }\ln . \\ x=\frac{1}{6}{e}^{(\frac{7}{2})} & \text{Divide by 6}.\end{array}\]

Example

Try it.

Solve \(\ln x=3.\)

Solution

\[\begin{array}{ll}\ln x=3 & \\ x={e}^{3} & \text{Use the definition of the natural logarithm}\text{.}\end{array}\]

represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words \({e}^{3}\approx 20.\) A calculator gives a better approximation: \({e}^{3}\approx 20.0855.\)

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0,\) \(S>0,\) \(T>0\) and any positive real number \(b,\) where \(b\ne 1,\)

\[{\log }_{b}S={\log }_{b}T\ \text{if and only if }S=T.\]

For example,

\[\text{If }{\log }_{2}(x-1)={\log }_{2}(8),\text{then }x-1=8.\]

So, if \(x-1=8,\) then we can solve for \(x,\) and we get \(x=9.\) To check, we can substitute \(x=9\) into the original equation: \({\log }_{2}(9-1)={\log }_{2}(8)=3.\) In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation \(\log (3x-2)-\log (2)=\log (x+4).\) To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x:\)

\[\begin{array}{ll}\log (3x-2)-\log (2)=\log (x+4) & \\ \log (\frac{3x-2}{2})=\log (x+4) & \text{Apply the quotient rule of logarithms}. \\ \frac{3x-2}{2}=x+4 & \text{Apply the one to one property of a logarithm}. \\ 3x-2=2x+8 & \text{Multiply both sides of the equation by }2. \\ x=10 & \text{Subtract 2}x\ \text{and add 2}.\end{array}\]

To check the result, substitute \(x=10\) into \(\log (3x-2)-\log (2)=\log (x+4).\)

\[\begin{array}{ll}\log (3(10)-2)-\log (2)=\log ((10)+4) & \\ \log (28)-\log (2)=\log (14) & \\ \log (\frac{28}{2})=\log (14) & \text{The solution checks}.\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Applied Problems Using Exponential and Logarithmic Equations

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. lists the half-life for several of the more common radioactive substances.

Substance	Use	Half-life
gallium-67	nuclear medicine	80 hours
cobalt-60	manufacturing	5.3 years
technetium-99m	nuclear medicine	6 hours
americium-241	construction	432 years
carbon-14	archeological dating	5,730 years
uranium-235	atomic power	703,800,000 years

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

\[\begin{array}{l}A(t)={A}_{0}{e}^{\frac{\ln (0.5)}{T}t} \\ A(t)={A}_{0}{e}^{\ln (0.5)\frac{t}{T}} \\ A(t)={A}_{0}{({e}^{\ln (0.5)})}^{\frac{t}{T}} \\ A(t)={A}_{0}{(\frac{1}{2})}^{\frac{t}{T}}\end{array}\]

where

\({A}_{0}\) is the amount initially present
\(T\) is the half-life of the substance
\(t\) is the time period over which the substance is studied
\(A(t)\) is the amount of the substance present after time \(t\)

Example

Try it.

How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Solution

\[\begin{array}{lllll}y=\text{1000}e\frac{\ln (0.5)}{\text{703,800,000}}t & \\ 900=1000{e}^{\frac{\ln (0.5)}{\text{703,800,000}}t} & \text{After 10\% decays, 900 grams are left}. \\ 0.9={e}^{\frac{\ln (0.5)}{\text{703,800,000}}t} & \text{Divide by 1000}. \\ \ln (0.9)=\ln ({e}^{\frac{\ln (0.5)}{\text{703,800,000}}t}) & \text{Take ln of both sides}. \\ \ln (0.9)=\frac{\ln (0.5)}{\text{703,800,000}}t & \text{ln}({e}^{M})=M \\ t=\text{703,800,000}\times \frac{\ln (0.9)}{\ln (0.5)}\text{years}\begin{array}{llll} & & & \end{array} & \text{Solve for}\ t. \\ t\approx \text{106,979,777 years} & \end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

One-to-one property for exponential functions	For any algebraic expressions \(\ S\) and \(\ T\) and any positive real number \(\ b,\) where \({b}^{S}={b}^{T}\) if and only if \(\ S=T.\)
Definition of a logarithm	For any algebraic expression S and positive real numbers \(\ b\) and \(\ c,\) where \(\ b\ne 1,\) \({\log }_{b}(S)=c\) if and only if \(\ {b}^{c}=S.\)
One-to-one property for logarithmic functions	For any algebraic expressions S and T and any positive real number \(\ b,\) where \(\ b\ne 1,\) \({\log }_{b}S={\log }_{b}T\) if and only if \(\ S=T.\)

Key Concepts

We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See .
When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See , , and .
When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See .
We can solve exponential equations with base \(e,\) by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See and .
After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See .
When given an equation of the form \({\log }_{b}(S)=c,\) where \(S\) is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation \({b}^{c}=S,\) and solve for the unknown. See and .
We can also use graphing to solve equations with the form \({\log }_{b}(S)=c.\) We graph both equations \(y={\log }_{b}(S)\) and \(y=c\) on the same coordinate plane and identify the solution as the x-value of the intersecting point. See .
When given an equation of the form \({\log }_{b}S={\log }_{b}T,\) where \(S\) and \(T\) are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation \(S=T\) for the unknown. See .
Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \({3}^{2x-5}=27.\)

Откриј одговор.

Is here a common base?	Yes, both 3 and 27 can be rewritten as powers of 3.
Write both sides of the equation with the same base.	\({3}^{2x-5}={3}^{3}\)
Since the bases are the same, the exponents must be equal.
Write a new equation by setting the exponents equal.	\(2x-5=3\)
Solve the equation.	\(2x=8\)
	\(x=4\)
Check the solution by substituting x=4 into the original equation.	\({3}^{2(4)-5}=27\)
	\(27=27,\) True

Solve \(3{e}^{x+2}=24\) . Find the exact answer and then approximate it to three decimal places.

Откриј одговор.

Rewriting with a common base is not possible.
Isolate the exponential by dividing both sides by 3.	\({e}^{x+2}=8\)
Take the natural logarithm of both sides.	\(\ln {e}^{x+2}=\ln 8\)
Use the Power Property to get the x as a factor, not an exponent.	\((x+2)\ln e=\ln 8\)
Use the property \(\ln e=1\) to simplify.	\(x+2=\ln 8\)
Solve the equation. Find the exact answer.	\(x=\ln 8-2\)
Approximate the answer.	\(x=0.079\)

Solve \(2({5}^{x})=12\) .
Isolate the exponential term on one side.
Take ln or log of each side.
Use the Power Property to get the x as a factor, not an exponent.
Solve for x. Give an exact answer and approximate. Check.
Solve \({2}^{3x-4}={8}^{-x}\).
Is here a common base here? Yes, both 2 and 8 can be rewritten as powers of 2.
Rewrite each side with a base of 2 using properties of exponents.
Set exponents equal since the bases are the same.
Solve for x. Give an exact answer and approximate. Check.
\({4}^{2x-3}=\frac{1}{16}\)
\(5({3}^{x})=20\)
Solve: \({\log }_{2}(3x-5)=4\)

Откриј одговор.

Rewrite in exponential form. \({2}^{4}=3x-5\)
Simplify. \(16=3x-5\)
Solve for x. \(x=7\)
Check. \({\log }_{2}(3(7)-5)=4\)
\(4=4\), True

Solve \({\log }_{4}(x+6)-{\log }_{4}(2x+5)=-{\log }_{4}x\)

Откриј одговор.

Use the Quotient Property on the left side and the Power Property on the right.	\({\log }_{4}\frac{x+6}{2x+5}={\log }_{4}{x}^{-1}\)
Rewrite \({x}^{-1}\) as \(\frac{1}{x}\).	\({\log }_{4}\frac{x+6}{2x+5}={\log }_{4}\frac{1}{x}\)
Use the One-to-One Property.	\(\frac{x+6}{2x+5}=\frac{1}{x}\)
Solve the rational equation.	\(x(x+6)=2x+5\)
Distribute and write in standard form.	\({x}^{2}+4x-5=0\)
Factor and solve for x.	\((x+5)(x-1)=0\) , \(x=-5\) , \(x=1\)
Check:	x=–5 is extraneous solution because \(2(-5)+5<0\) so x=1 is the only solution.

Use the following steps to help solve the equation below.
Solve \(\log (x+2)-\log 3=1\)
Use properties of logarithms to rewrite the left side as a single log term.
Convert to exponential form.
Solve for x. Check.
Use the following steps to help solve the equation below.
Solve \(\log x+\log (x+1)=2\)
Use properties of logarithms to rewrite the left side as a single log term.
Use the One-to-One Property.
Solve the quadratic equation.
Check.
\({\log }_{3}x=5\)
\({\log }_{2}(x+1)+{\log }_{2}(x-1)=3\)
\(\log (x-2)-\log (4x+16)=\log \frac{1}{x}\)
Solve \({2}^{x-1}={2}^{2x-4}.\)

Откриј одговор.

\[\begin{array}{lllll}{2}^{x-1}={2}^{2x-4} & \text{The common base is }2. \\ x-1=2x-4\begin{array}{llll} & & & \end{array} & \text{By the one-to-one property the exponents must be equal}. \\ x=3 & \text{Solve for }x.\end{array}\]
Solve \({5}^{2x}={5}^{3x+2}.\)

Откриј одговор.

\(x=-2\)
Solve \({8}^{x+2}={16}^{x+1}.\)

Откриј одговор.

\[\begin{array}{ll}{8}^{x+2}={16}^{x+1} & \\ {({2}^{3})}^{x+2}={({2}^{4})}^{x+1} & \text{Write}\ 8\ \text{and}\ 16\ \text{as powers of}\ 2. \\ {2}^{3x+6}={2}^{4x+4} & \text{To take a power of a power, multiply exponents}. \\ 3x+6=4x+4 & \text{Use the one-to-one property to set the exponents equal}. \\ x=2 & \text{Solve for }x.\end{array}\]
Solve \({5}^{2x}={25}^{3x+2}.\)

Откриј одговор.

\(x=-1\)
Solve \({2}^{5x}=\sqrt{2}.\)

Откриј одговор.

\[\begin{array}{ll}{2}^{5x}={2}^{\frac{1}{2}} & \text{Write the square root of 2 as a power of}\ 2. \\ 5x=\frac{1}{2} & \text{Use the one-to-one property}. \\ x=\frac{1}{10} & \text{Solve for}\ x.\end{array}\]
Solve \({5}^{x}=\sqrt{5}.\)

Откриј одговор.

\(x=\frac{1}{2}\)
Solve \({3}^{x+1}=-2.\)

Откриј одговор.

This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.
Solve \({2}^{x}=-100.\)

Откриј одговор.

The equation has no solution.
Solve \({5}^{x+2}={4}^{x}.\)

Откриј одговор.

\[\begin{array}{lllll}{5}^{x+2}={4}^{x} & \text{There is no easy way to get the powers to have the same base}. \\ \ln {5}^{x+2}=\ln {4}^{x} & \text{Take ln of both sides}. \\ (x+2)\ln 5=x\ln 4 & \text{Use laws of logs}. \\ x\ln 5+2\ln 5=x\ln 4 & \text{Use the distributive law}. \\ x\ln 5-x\ln 4=-2\ln 5 & \text{Get terms containing}\ x\ \text{on one side, terms without}\ x\ \text{on the other}. \\ x(\ln 5-\ln 4)=-2\ln 5 & \text{On the left hand side, factor out an }x. \\ x\ln (\frac{5}{4})=\ln (\frac{1}{25})\begin{array}{llll} & & & \end{array} & \text{Use the laws of logs.} \\ x=\frac{\ln (\frac{1}{25})}{\ln (\frac{5}{4})} & \text{Divide by the coefficient of}\ x.\end{array}\]
Solve \({2}^{x}={3}^{x+1}.\)

Откриј одговор.

\(x=\frac{\ln 3}{\ln (\frac{2}{3})}\)
Solve \(100=20{e}^{2t}.\)

Откриј одговор.

\[\begin{array}{lll}100 & =20{e}^{2t} & \\ 5 & ={e}^{2t} & \text{Divide by the coefficient of the power.} \\ \ln 5 & =2t & \text{Take ln of both sides}\text{. Use the fact that }\ln (x)\ \text{and }{e}^{x}\ \text{are inverse functions}\text{.} \\ t & =\frac{\ln 5}{2} & \text{Divide by the coefficient of }t\text{.}\end{array}\]
Solve \(3{e}^{0.5t}=11.\)

Откриј одговор.

\(t=2\ln (\frac{11}{3})\) or \(\ln {(\frac{11}{3})}^{2}\)
Solve \(4{e}^{2x}+5=12.\)

Откриј одговор.

\[\begin{array}{ll}4{e}^{2x}+5=12 & \\ 4{e}^{2x}=7 & \text{Combine like terms}. \\ {e}^{2x}=\frac{7}{4} & \text{Divide by the coefficient of the power}. \\ 2x=\ln (\frac{7}{4}) & \text{Take ln of both sides}. \\ x=\frac{1}{2}\ln (\frac{7}{4}) & \text{Solve for }x.\end{array}\]
Solve \(3+{e}^{2t}=7{e}^{2t}.\)

Откриј одговор.

\(t=\ln (\frac{1}{\sqrt{2}})=-\frac{1}{2}\ln (2)\)
Solve \({e}^{2x}-{e}^{x}=56.\)

Откриј одговор.

\[\begin{array}{llllll}{e}^{2x}-{e}^{x} & =56 & \\ {e}^{2x}-{e}^{x}-56 & =0 & \text{Get one side of the equation equal to zero}. \\ ({e}^{x}+7)({e}^{x}-8) & =0 & \text{Factor by the FOIL method}. \\ {e}^{x}+7 & =0\ \text{or }{e}^{x}-8=0\begin{array}{llll} & & & \end{array} & \text{If a product is zero, then one factor must be zero}. \\ {e}^{x} & =-7{\ \text{or e}}^{x}=8 & \text{Isolate the exponentials}. \\ {e}^{x} & =8 & \text{Reject the equation in which the power equals a negative number}. \\ x & =\ln 8 & \text{Solve the equation in which the power equals a positive number}.\end{array}\]
Solve \({e}^{2x}={e}^{x}+2.\)

Откриј одговор.

\(x=\ln 2\)
Solve \(2\ln x+3=7.\)

Откриј одговор.

\[\begin{array}{ll}2\ln x+3=7 & \\ 2\ln x=4 & \text{Subtract 3}. \\ \ln x=2 & \text{Divide by 2}. \\ x={e}^{2} & \text{Rewrite in exponential form}.\end{array}\]
Solve \(6+\ln x=10.\)

Откриј одговор.

\(x={e}^{4}\)
Solve \(2\ln (6x)=7.\)

Откриј одговор.

\[\begin{array}{ll}2\ln (6x)=7 & \\ \ln (6x)=\frac{7}{2} & \text{Divide by 2}. \\ 6x={e}^{(\frac{7}{2})} & \text{Use the definition of }\ln . \\ x=\frac{1}{6}{e}^{(\frac{7}{2})} & \text{Divide by 6}.\end{array}\]
Solve \(2\ln (x+1)=10.\)

Откриј одговор.

\(x={e}^{5}-1\)
Solve \(\ln x=3.\)

Откриј одговор.

\[\begin{array}{ll}\ln x=3 & \\ x={e}^{3} & \text{Use the definition of the natural logarithm}\text{.}\end{array}\]
represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words \({e}^{3}\approx 20.\) A calculator gives a better approximation: \({e}^{3}\approx 20.0855.\)
Use a graphing calculator to estimate the approximate solution to the logarithmic equation \({2}^{x}=1000\) to 2 decimal places.

Откриј одговор.

\(x\approx 9.97\)
Solve \(\ln ({x}^{2})=\ln (2x+3).\)

Откриј одговор.

\[\begin{array}{lllll}\ln ({x}^{2})=\ln (2x+3) & \\ {x}^{2}=2x+3 & \text{Use the one-to-one property of the logarithm}. \\ {x}^{2}-2x-3=0 & \text{Get zero on one side before factoring}. \\ (x-3)(x+1)=0 & \text{Factor using FOIL}. \\ x-3=0\ \text{or }x+1=0\begin{array}{llll} & & & \end{array} & \text{If a product is zero, one of the factors must be zero}. \\ x=3\ \text{or}\ x=-1 & \text{Solve for }x.\end{array}\]
Solve \(\ln ({x}^{2})=\ln 1.\)

Откриј одговор.

\(x=1\) or \(x=-1\)
How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Откриј одговор.

\[\begin{array}{lllll}y=\text{1000}e\frac{\ln (0.5)}{\text{703,800,000}}t & \\ 900=1000{e}^{\frac{\ln (0.5)}{\text{703,800,000}}t} & \text{After 10\% decays, 900 grams are left}. \\ 0.9={e}^{\frac{\ln (0.5)}{\text{703,800,000}}t} & \text{Divide by 1000}. \\ \ln (0.9)=\ln ({e}^{\frac{\ln (0.5)}{\text{703,800,000}}t}) & \text{Take ln of both sides}. \\ \ln (0.9)=\frac{\ln (0.5)}{\text{703,800,000}}t & \text{ln}({e}^{M})=M \\ t=\text{703,800,000}\times \frac{\ln (0.9)}{\ln (0.5)}\text{years}\begin{array}{llll} & & & \end{array} & \text{Solve for}\ t. \\ t\approx \text{106,979,777 years} & \end{array}\]
How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

Откриј одговор.

\(t=703,800,000\times \frac{\ln (0.8)}{\ln (0.5)}\ \text{years }\approx \ 226,572,993\ \text{years}.\)
How can an exponential equation be solved?

Откриј одговор.

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Exponential and Logarithmic Equations

Use like bases to solve exponential equations.
Use logarithms to solve exponential equations.
Use the definition of a logarithm to solve logarithmic equations.
Use the one-to-one property of logarithms to solve logarithmic equations.
Solve applied problems involving exponential and logarithmic equations.
Solve Exponential Equations. (IA 10.2.2)
Solve Logarithmic Equations. (IA 10.3.4)
Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Више у Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Rewrite in exponential form.	\({2}^{4}=3x-5\)
Simplify.	\(16=3x-5\)
Solve for x.	\(x=7\)
Check.	\({\log }_{2}(3(7)-5)=4\)
	\(4=4\), True