# 6.6 Exponential and logarithmic equations  (Page 3/8)

 Page 3 / 8

Solve $\text{\hspace{0.17em}}{2}^{x}={3}^{x+1}.$

$x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(2}{3}\right)}$

Is there any way to solve $\text{\hspace{0.17em}}{2}^{x}={3}^{x}?$

Yes. The solution is $0.$

## Equations containing e

One common type of exponential equations are those with base $\text{\hspace{0.17em}}e.\text{\hspace{0.17em}}$ This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ on either side, we can use the natural logarithm    to solve it.

Given an equation of the form $\text{\hspace{0.17em}}y=A{e}^{kt}\text{,}$ solve for $\text{\hspace{0.17em}}t.$

1. Divide both sides of the equation by $\text{\hspace{0.17em}}A.$
2. Apply the natural logarithm of both sides of the equation.
3. Divide both sides of the equation by $\text{\hspace{0.17em}}k.$

## Solve an equation of the form y = Ae kt

Solve $\text{\hspace{0.17em}}100=20{e}^{2t}.$

Solve $\text{\hspace{0.17em}}3{e}^{0.5t}=11.$

$t=2\mathrm{ln}\left(\frac{11}{3}\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}$

Does every equation of the form $\text{\hspace{0.17em}}y=A{e}^{kt}\text{\hspace{0.17em}}$ have a solution?

No. There is a solution when $\text{\hspace{0.17em}}k\ne 0,$ and when $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is $\text{\hspace{0.17em}}2=-3{e}^{t}.$

## Solving an equation that can be simplified to the form y = Ae kt

Solve $\text{\hspace{0.17em}}4{e}^{2x}+5=12.$

Solve $\text{\hspace{0.17em}}3+{e}^{2t}=7{e}^{2t}.$

$t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)$

## Extraneous solutions

Sometimes the methods used to solve an equation introduce an extraneous solution    , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

## Solving exponential functions in quadratic form

Solve $\text{\hspace{0.17em}}{e}^{2x}-{e}^{x}=56.$

Solve $\text{\hspace{0.17em}}{e}^{2x}={e}^{x}+2.$

$x=\mathrm{ln}2$

Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

## Using the definition of a logarithm to solve logarithmic equations

We have already seen that every logarithmic equation $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right)=y\text{\hspace{0.17em}}$ is equivalent to the exponential equation $\text{\hspace{0.17em}}{b}^{y}=x.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x-5\right)=3.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x:$

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