# Question #1b9c7

Apr 5, 2017

Solution to a) $\to x = \frac{1}{2}$

#### Explanation:

Solution to a)

Adding logs reflects the action of multiplication of the source values.

So $\ln \left(x - 1\right) + \ln \left(3\right) = \ln \left(x\right)$ is the same as: $\ln \left(3 \left[x - 1\right]\right) = \ln \left(x\right)$

Using an example: note that ${\ln}^{- 1} \left(a\right) = a$

So $\text{ "ln^(-1)(3[x-1])" "=" } {\ln}^{- 1} \left(x\right)$

$\text{ "3(x-1)" "=" } x$

$\text{ } 3 x - 1 = x$

$\text{ } 2 x = 1$

$\text{ } x = \frac{1}{2}$

Apr 5, 2017

Solution to b) $\text{ } x \approx 8.773 \ldots$ to 3 decimal places

#### Explanation:

Solution to b)

Using an example: note that $\text{ } 2 \ln \left(a\right) \to \ln \left({a}^{2}\right)$

So we have:

$\ln \left[{\left(5 x\right)}^{3}\right] - \ln \left({x}^{2}\right) = 7$

$\ln \left[\frac{125 {x}^{3}}{x} ^ 2\right] = 7$

$\ln \left(125 x\right) = 7$

${\ln}^{- 1} \left(125 x\right) = {\ln}^{- 1} \left(7\right) \leftarrow \text{ not convinced this will work!}$

Trying something else!

$\ln \left(125\right) + \ln \left(x\right) = 7$

$\ln \left(x\right) = 7 - \ln \left(125\right)$

$x = {\ln}^{- 1} \left(x\right) = {\ln}^{- 1} \left[\textcolor{w h i t e}{\frac{2}{2}} 7 - \ln \left(125\right) \textcolor{w h i t e}{\frac{2}{2}}\right]$

$x \approx 8.773 \ldots$ to 3 decimal places