# Question #3a27d

Apr 8, 2017

#### Explanation:

Given: $77 = {33}^{x}$

Use the natural logarithm on both sides:

$\ln \left(77\right) = \ln \left({33}^{x}\right)$

All logarithms have the property $\ln \left({a}^{c}\right) = \left(c\right) \ln \left(a\right)$; we shall use the property on the right side:

$\ln \left(77\right) = \left(x\right) \ln \left(33\right)$

Divide both sides by $\ln \left(33\right)$:

$x = \ln \frac{77}{\ln} \left(33\right)$

The following is an approximation rounded to seven decimal places:

$x \approx 1.2423269$

Given: $0.77 = \ln \left(y\right)$

Flip the equation:

$\ln \left(y\right) = 0.77$

Make both sides an exponent of the exponential function:

${e}^{\ln \left(y\right)} = {e}^{0.77}$

We do this, because the exponential function and the natural logarithm cancel each other, thereby, leaving only y on the left:

$y = {e}^{0.77}$

The following is an approximation rounded to seven decimal places:

$y \approx 2.1597663$