1. ## 3 more problems

1. 2ln(3x) = 8

2. Logx + log (x + 15) = 2

3. Ln (x - 4) - ln (x + 1) = ln 6

2. Originally Posted by garbles
1. 2ln(3x) = 8
$2\ln \left( {3x} \right) = 8 \Leftrightarrow \ln \left( {3x} \right) = 4 \Leftrightarrow 3x = {e^4} \Leftrightarrow x = \frac{{{e^4}}}{3}.$

3. Thanks just need the last 2

4. Originally Posted by garbles

3. Ln (x - 4) - ln (x + 1) = ln 6
$\underbrace {\ln \left( {x - 4} \right) - \ln \left( {x + 1} \right)}_{\ln a - \ln b = \ln \frac{a}{b}} = \ln 6 \Leftrightarrow \ln \frac{{x - 4}}{{x + 1}} = \ln 6 \Leftrightarrow$

$\Leftrightarrow \frac{{x - 4}}{{x + 1}} = 6 \Leftrightarrow x - 4 = 6\left( {x + 1} \right) \Leftrightarrow 5x = - 10 \Leftrightarrow x = - 2.$

5. Originally Posted by garbles

2. Logx + log (x + 15) = 2
What is a basis of these logarithms?

${\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.$

6. Originally Posted by DeMath
What is a basis these logarithms?

${\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.$
It doesn't say

7. Originally Posted by DeMath
What is a basis these logarithms?

${\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.$
I'd just assume base 10.

So, $log_{10}(x(x+15))=log_{10}(x^{2}+15x)$

Thus, $10^{2}=x^{2}+15x$ and $x^{2}+15x-100=0$ becomes the equation you have to solve for.

You will want to keep the non-negative answer.

8. Originally Posted by garbles
2. $\Log{x}+ \log {(x + 15)} = 2$

$\log{x}+\log{(x+15)}=2$
$\log{(x(x+15))}=2$
$\log{(x^2+15x)}=2$
$x^2+15x=10^2$
$x^2+15x-100=0$
$(x+20)(x-5)=0$
x=5 not -20 since $\log{-20}$ is undefined