Question #3a27d

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Answer 1 · 2017-04-08T02:45:22

Please see the explanation

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

Use the natural logarithm on both sides:

$ln (77) = ln (33^{x})$ $ln(77)=ln(33^x)$

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

$ln (77) = (x) ln (33)$ $ln(77)=(x)ln(33)$

Divide both sides by $ln (33)$ $ln(33)$ :

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

The following is an approximation rounded to seven decimal places:

$x \approx 1.2423269$ $x ~~ 1.2423269$

Given: $0.77 = ln (y)$ $0.77=ln(y)$

Flip the equation:

$ln (y) = 0.77$ $ln(y) = 0.77$

Make both sides an exponent of the exponential function:

$e^{ln (y)} = e^{0.77}$ $e^(ln(y)) = e^0.77$

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

$y = e^{0.77}$ $y = e^0.77$

The following is an approximation rounded to seven decimal places:

$y \approx 2.1597663$ $y ~~ 2.1597663$