how do i write log base 8 of x as a log base 2 of x??

8=2^3
people are telling you to use the change of base formula, log_a(x) = log_b(x)/log_b(a). that works, but it's not a good idea to rely on formulas like that that you probably don't understand

all you need to do is remember what log means. log_2(x) means the power that you need to raise 2 to so that the result is x. log_8(x) means the power that you need to raise 8 to so that the result is x. now think about what raising 8 to a power means. it just means multiplying lots of copies of 8 together^(). but because 8 is 2\*2\*2, multiplying n copies of 8 together is the same as multiplying 3\*n copies of 2 together. 8^n = 8\*8\*...\*8 = (2\*2\*2)\*(2\*2\*2)\*...\*(2\*2\*2) = 2^(3n).

so suppose we raise 8 to the power of something and the result is x (remember that this "something" is what we call log_8(x)). if we want to raise 2 to the power of something else and get x (this "something else" is log_2(x)), we need to have three times as many 2s as we have 8s. so, number of 2s to multiply = 3 * number of 8s to multiply. or in other words, log_2(x) = 3 log_8(x).

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: at least if the exponent is a positive integer, but most of the formulas that you can find by thinking about exponentiation as repeated multiplication will still be true more generally.
log\_8 x = (ln x)/(ln 8) = (ln x)/(3 ln 2) = (1/3)(ln x)/(ln 2) = (1/3)log\_2 x
log base b^n of x = log base b of the nth_root of x.

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