If $(1 + x)^{10} = 2.5$, what is $x$? If
$$(1 + x)^{10} = 2.5$$
, the next step would be
$$10 * \log(1 + x) = \log(2.5)$$
$$\log(1 + x) = \frac{\log(2.5)}{10}$$
. This is where I am stuck. I know that
$$\log_{b}(a + c) = \log_{b}(a) + \log_b\left( 1 + \frac{c}{a} \right)$$
, but that does not help me in any way.
 A: When the unknown isn't the exponent, but the base, then logarithms aren't the most natural solution. Roots are:
$$
(1+x)^{10} = 2.5\\
1+x = \pm\sqrt[10]{2.5}\\
x = -1\pm\sqrt[10]{2.5}
$$
A: Just, $$x=2.5^{0.1}-1$$ or $$x=-2.5^{0.1}-1.$$
A: $(1+x)^{10} = 2.5$                  (original equation)
$1+x = \sqrt[10]{2.5} $            (take 10th root of both sides)
$1+x = \pm1.09596... $             (simplify)
$x = 0.09596...$                      (subtract 1 on both sides, positive root)
or
$x = -2.09596... $                    (subtract 1 on both sides, negative root)
A: If the equation $\log(1+x) = \frac{\log(2.5)}{10}$ holds then this equation holds no matter what base the logs were taken in (as long as the logs on both sides were taken to same base). Make sure you see why.
Calculate $a=\log_e(2.5)$. Then $1+x = e^{a/10}$ which gives $x= e^{a/10}-1$.
If you are more comfortable working base 10, then
$a=\log_{10}(2.5)$. Then $1+x = 10^{a/10}$ which gives $x= 10^{a/10}-1$.

ETA: you should check that this answer and Arthur's above give the same result.
