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?
Remember that you can remove the bases once there are only logs. So, raise the power
$\displaystyle 2log_2(x+2) = \log_2(x+14)$
$\displaystyle log_2(x+2)^2 = \log_2(x+14)$
$\displaystyle (x+2)^2 = x + 14$
For the other one, change the base.
$\displaystyle log_a b = \dfrac{1}{\log_b a}$
EDIT: I didn't notice the '2'. Changed it now.
$\displaystyle 2\log_2(x+2) = \log_2[(x+2)^2]$
In other words $\displaystyle (x+2)^2 = x+14$ - solve the quadratic but remember that x must be greater than -2
For number 2 use the change of base rule
$\displaystyle \dfrac{\ln(x)}{\ln(27)} = \dfrac{\ln(27)}{\ln(x)} \rightarrow \: [\ln(x)]^2 - [\ln(27)]^2 = 0$
Solve using the difference of two squares. Again, from the domain, $\displaystyle x > 0 \text { and } x \neq 1$
$\displaystyle \dfrac{1}{\log_x(27)} = \log_x (27) $
Yes that is correct.
Now bring the x's to one side giving us:
$\displaystyle 1 = (\log_x (27))^2$
$\displaystyle \sqrt{1} = \log_x (27)$
Can you carry on from there?
(Remember $\displaystyle \sqrt{1}$ gives 2 answers)