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I have the following formula:

$$a \times (1+z)^p = b$$

How can I solve for $z$?

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    You need to first list some assumptions on what $a,b,p$ are, then decide if you look for real vs. complex solutions, at the very least. – 2017-02-28

2 Answers 2

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Very little effort is required to show that: \begin{align} (1 + z)^p &= \frac{b}{a} \\ 1 + z &= \left(\frac{b}{a}\right)^{1/p} \\ z &= \left(\frac{b}{a}\right)^{1/p} -1 \end{align}

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    Couldn't have had a better answer. Thanks a lot. Looks so easy that my initial question almost appears dumb... – 2017-02-28
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This is not a polynomial equation. You need to remember enough about the laws of exponents to solve it.

$$ a(1+z)^p = b $$ implies $$ 1+z = (b/a)^{1/p} $$ so $$ z = (b/a)^{1/p} -1 \ $$