Question

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

Answer 1

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!