How do you solve log_2 x - log_8 x = 4?

1 Answer
Apr 7, 2016

You must first put in the same base. This can be started out by using the log rule log_an = logn/loga

Explanation:

logx/log2 - logx/log8 = 4

Now, rewrite the terms in the denominator by using the rule loga^n = nloga

logx/(1log2) - logx/(log2^3) = 4

logx/(1log2) - logx/(3log2) = 4

Place on an equal denominator.

(3logx)/(3log2) - logx/(3log2) = 4

log_8(x^3) - log_8(x) = 4

Now, you must use the rule log_an - log_am = log_a(n/m).

log_8(x^3/x) = 4

Convert to exponential form:

x^2 = 8^4

x^2 = 4096

x = 64

Checking the solution in the equation, we find that it works. Thus our solution set is {x = 64}

Hopefully this helps!