Finite Math Examples

$2 p = p {(1 + \frac{0.2152}{1})}^{t}$

Step 1

Rewrite the equation as $p {(1 + \frac{0.2152}{1})}^{t} = 2 p$ .

$p {(1 + \frac{0.2152}{1})}^{t} = 2 p$

Step 2

Simplify.

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Step 2.1

Divide $0.2152$ by $1$ .

$p {(1 + 0.2152)}^{t} = 2 p$

Step 2.2

Add $1$ and $0.2152$ .

$p \cdot {1.2152}^{t} = 2 p$

Step 3

Divide each term in $p \cdot {1.2152}^{t} = 2 p$ by $p$ and simplify.

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Step 3.1

Divide each term in $p \cdot {1.2152}^{t} = 2 p$ by $p$ .

$\frac{p \cdot {1.2152}^{t}}{p} = \frac{2 p}{p}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $p$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{p \cdot {1.2152}^{t}}{p} = \frac{2 p}{p}$

Step 3.2.1.2

Divide ${1.2152}^{t}$ by $1$ .

${1.2152}^{t} = \frac{2 p}{p}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $p$ .

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Step 3.3.1.1

Cancel the common factor.

${1.2152}^{t} = \frac{2 p}{p}$

Step 3.3.1.2

Divide $2$ by $1$ .

${1.2152}^{t} = 2$

Step 4

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

$\ln ({1.2152}^{t}) = \ln (2)$

Step 5

Expand $\ln ({1.2152}^{t})$ by moving $t$ outside the logarithm.

$t \ln (1.2152) = \ln (2)$

Step 6

Divide each term in $t \ln (1.2152) = \ln (2)$ by $\ln (1.2152)$ and simplify.

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Step 6.1

Divide each term in $t \ln (1.2152) = \ln (2)$ by $\ln (1.2152)$ .

$\frac{t \ln (1.2152)}{\ln (1.2152)} = \frac{\ln (2)}{\ln (1.2152)}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $\ln (1.2152)$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{t \ln (1.2152)}{\ln (1.2152)} = \frac{\ln (2)}{\ln (1.2152)}$

Step 6.2.1.2

Divide $t$ by $1$ .

$t = \frac{\ln (2)}{\ln (1.2152)}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$t = \frac{\ln (2)}{\ln (1.2152)}$

Decimal Form:

$t = 3.55626649 \dots$