Precalculus Examples

$4 e^{6 t} = 6 e^{5 t}$

Step 1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (4 e^{6 t}) = \ln (6 e^{5 t})$

Step 2

Expand the left side.

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Step 2.1

Rewrite $\ln (4 e^{6 t})$ as $\ln (4) + \ln (e^{6 t})$ .

$\ln (4) + \ln (e^{6 t}) = \ln (6 e^{5 t})$

Step 2.2

Expand $\ln (e^{6 t})$ by moving $6 t$ outside the logarithm.

$\ln (4) + 6 t \ln (e) = \ln (6 e^{5 t})$

Step 2.3

The natural logarithm of $e$ is $1$ .

$\ln (4) + 6 t \cdot 1 = \ln (6 e^{5 t})$

Step 2.4

Multiply $6$ by $1$ .

$\ln (4) + 6 t = \ln (6 e^{5 t})$

Step 3

Expand the right side.

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Step 3.1

Rewrite $\ln (6 e^{5 t})$ as $\ln (6) + \ln (e^{5 t})$ .

$\ln (4) + 6 t = \ln (6) + \ln (e^{5 t})$

Step 3.2

Expand $\ln (e^{5 t})$ by moving $5 t$ outside the logarithm.

$\ln (4) + 6 t = \ln (6) + 5 t \ln (e)$

Step 3.3

The natural logarithm of $e$ is $1$ .

$\ln (4) + 6 t = \ln (6) + 5 t \cdot 1$

Step 3.4

Multiply $5$ by $1$ .

$\ln (4) + 6 t = \ln (6) + 5 t$

Step 4

Move all the terms containing a logarithm to the left side of the equation.

$\ln (4) - \ln (6) = - 6 t + 5 t$

Step 5

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{4}{6}) = - 6 t + 5 t$

Step 6

Cancel the common factor of $4$ and $6$ .

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Step 6.1

Factor $2$ out of $4$ .

$\ln (\frac{2 (2)}{6}) = - 6 t + 5 t$

Step 6.2

Cancel the common factors.

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Step 6.2.1

Factor $2$ out of $6$ .

$\ln (\frac{2 \cdot 2}{2 \cdot 3}) = - 6 t + 5 t$

Step 6.2.2

Cancel the common factor.

$\ln (\frac{2 \cdot 2}{2 \cdot 3}) = - 6 t + 5 t$

Step 6.2.3

Rewrite the expression.

$\ln (\frac{2}{3}) = - 6 t + 5 t$

Step 7

Add $- 6 t$ and $5 t$ .

$\ln (\frac{2}{3}) = - t$

Step 8

Since $t$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- t = \ln (\frac{2}{3})$

Step 9

Divide each term in $- t = \ln (\frac{2}{3})$ by $- 1$ and simplify.

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Step 9.1

Divide each term in $- t = \ln (\frac{2}{3})$ by $- 1$ .

$\frac{- t}{- 1} = \frac{\ln (\frac{2}{3})}{- 1}$

Step 9.2

Simplify the left side.

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Step 9.2.1

Dividing two negative values results in a positive value.

$\frac{t}{1} = \frac{\ln (\frac{2}{3})}{- 1}$

Step 9.2.2

Divide $t$ by $1$ .

$t = \frac{\ln (\frac{2}{3})}{- 1}$

Step 9.3

Simplify the right side.

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Step 9.3.1

Move the negative one from the denominator of $\frac{\ln (\frac{2}{3})}{- 1}$ .

$t = - 1 \cdot \ln (\frac{2}{3})$

Step 9.3.2

Rewrite $- 1 \cdot \ln (\frac{2}{3})$ as $- \ln (\frac{2}{3})$ .

$t = - \ln (\frac{2}{3})$

Step 10

The result can be shown in multiple forms.

Exact Form:

$t = - \ln (\frac{2}{3})$

Decimal Form:

$t = 0.40546510 \dots$