# If y=ab^x what is x ?

## Dec 31, 2017

$x = \ln \frac{y \frac{\setminus}{a}}{\ln} \left(b\right)$

#### Explanation:

First we divide both sides by $a$ to get:
$\frac{y}{a} = {b}^{x}$

To crack this apart further, we could take ${\log}_{b}$ on both sides, but none of the alternatives involve ${\log}_{b}$. I will instead use $\ln$ on both sides and then take advantage of some log properties.
$\ln \left(\frac{y}{a}\right) = \ln \left({b}^{x}\right)$

Now we can use the logarithm properties to move the $x$ power out the front of the logarithm:
$\ln \left(\frac{y}{a}\right) = x \ln \left(b\right)$

Now we divide both sides by $\ln \left(b\right)$:
$\ln \frac{\frac{y}{a}}{\ln} \left(b\right) = x$

This is the same as alternative A, which is the correct answer.

Dec 31, 2017

C or may be A

#### Explanation:

we know that if $a = {b}^{c}$
then
$c = \ln \frac{a}{\ln} b$ similarly
if $y = a {b}^{x}$
then
$x = \ln \frac{y}{\ln} a b$

. but wait , there could be an alternate solution , may be it would be more correct than it is!
i assumed that the question was
${\left(a b\right)}^{x} = y$
it could be
$a \times {b}^{x} = y$
in this case
first divide both sides by a that makes
${b}^{x} = \frac{y}{a}$
on applying same principles we'd applied earlier
$\ln \frac{\frac{y}{a}}{\ln} b$ that's part A