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## anonymous 5 years ago log8(x+1)=2/3 (a small 8)

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1. anonymous

$\log_ab = c \iff a^c =b$

2. anonymous

$x+1=8^{\frac{2}{3}}$ $x+1=\sqrt[3]8^2$ $x+1=2^2$ $x+1=4$ $x=3$

3. anonymous

$\log_8(x+1)=\frac{2}{3}$ now we make both of the sides an exponent of 8. It'll look like this $8^{\log_8(x+1)}=8^{\frac{2}{3}}$ the left side simply becomes x+1 because of a common cancellation rule of logs. In general, $a^{\log_ab}=b$ since log base a of b is the exponent we must raise a to in order to get b, if we make log base a of b an exponent of a, we get b the left side is $8^\frac{2}{3}=\sqrt[3]{8}^2=2^2=4$ therefore, x+1=4, and x=3

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