solve for x: log base 8(x+1)=2/3

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solve for x: log base 8(x+1)=2/3

Mathematics
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so you'll rewrite that as (the third root of 8) or 2, then square that = x + 1 2^2 = 4 = x + 1 so x = 3
maybe i didnt write it right but its log small 8 (x+1)=2/3 can you explain how to solve that?
the (x+1) is not small right? and it's on the same line as Log?

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yeah only the 8 is smaller and a line below the log
Yeah I explained it above
do you move the 8 to the right and put it to the 2/3?
The formula is: [a ^{k} = x and that can be written as Log base a of x = k
i dont understand how to show my steps
In general, \[\log_{a}c = b \ \ , a^{b} = c \]
So your is Log base 8 of (x + 1) = 2/3 Therefore it's: 8^(2/3) = x + 1 8 ^ (2/3) = The third root of 8 ( or 2), then you square it ( or 4)! So you have 4 = x + 1 So x = 3 And yes, that's the same one except it's using a,b, and c instead! ^^
\[\log_{8} x+1 = \frac{2}{3} \ \ , 8^{\frac{2}{3}} = x+1\] \[8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\] \[4 = x+1 \ \ , x = 3\] Sorry if I'm invading, HanhLive; just formatting it so it's easier to read :)
okk thank you i got it

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