# Help With Logarithm Questions?

## Can you please solve for x and explain (show you work step by step) how you did questions 9, 10, 11, and 12. Please write your final answer in 3 significant figures.

May 3, 2017

Solution to question 9

$x \approx 4.42$ to 2 decimal places.

#### Explanation:

9) $\text{ } {2}^{x + 1} = {3}^{x - 1}$

Take logs of both sides

$\ln \left({2}^{x + 1}\right) = \ln \left({3}^{x - 1}\right)$

$\left(x + 1\right) \ln \left(2\right) = \left(x - 1\right) \ln \left(3\right)$

$x \ln \left(2\right) + \ln \left(2\right) = x \ln \left(3\right) - \ln \left(3\right)$

$x \ln \left(3\right) - x \ln \left(2\right) = \ln \left(2\right) + \ln \left(3\right)$

x=(ln(2)+ln(3))/(ln(3)-ln(2)

But:
Addition of logs is the consequence of multiplication of the source numbers.

Subtraction of logs is the consequence of division of the source numbers.

$x = \ln \frac{6}{\ln \left(\frac{3}{2}\right)}$

$x \approx 4.4190 \ldots .$

$x \approx 4.42$ to 2 decimal places.

May 3, 2017

Solution to question 10

$x \approx 321.70$ to 2 decimal places

$\textcolor{red}{\text{You should be able to do questions 11 and 12 now}}$

#### Explanation:

Note that ${\log}_{e} \left(e\right) \to \ln \left(e\right) = 1$ as does ${\log}_{10} \left(10\right) = 1$

$10 x \left[\textcolor{w h i t e}{.} \ln \left(2\right) + \ln \left(e\right) \textcolor{w h i t e}{.}\right] = 19$

$x = \frac{19}{10 \left(\ln \left(2\right) + 1\right)}$

$x \approx 321.6979 \ldots .$

$x \approx 321.70$ to 2 decimal places

May 3, 2017

$x = 0.22512918$

#### Explanation:

10) ${e}^{10 x} = \frac{19}{2} \mathmr{and} {e}^{10 x} = 9.5$ taking log in both sides,

10x*ln e =ln(9.5) or 10x =2.2512918 ; ln (e)=1:. x= 2.2512918/10=0.22512918 [Ans]