1. ## Logarithms

Solve each of the following eq.

1. 2 log(base 2)(x + 2) = log(base2)(x + 14)

2. log(base 27)(x) = log(base x)(27)

How to start?

2. Remember that you can remove the bases once there are only logs. So, raise the power

$2log_2(x+2) = \log_2(x+14)$

$log_2(x+2)^2 = \log_2(x+14)$

$(x+2)^2 = x + 14$

For the other one, change the base.

$log_a b = \dfrac{1}{\log_b a}$

EDIT: I didn't notice the '2'. Changed it now.

3. $2\log_2(x+2) = \log_2[(x+2)^2]$

In other words $(x+2)^2 = x+14$ - solve the quadratic but remember that x must be greater than -2

For number 2 use the change of base rule

$\dfrac{\ln(x)}{\ln(27)} = \dfrac{\ln(27)}{\ln(x)} \rightarrow \: [\ln(x)]^2 - [\ln(27)]^2 = 0$

Solve using the difference of two squares. Again, from the domain, $x > 0 \text { and } x \neq 1$

4. For no.2

After 1/log(base x)(27) =log(base x)(27)

Then,I'm kinda confused.....Is this the correct way?

5. $\dfrac{1}{\log_x(27)} = \log_x (27)$

Yes that is correct.

Now bring the x's to one side giving us:

$1 = (\log_x (27))^2$

$\sqrt{1} = \log_x (27)$

Can you carry on from there?
(Remember $\sqrt{1}$ gives 2 answers)