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

Question

6433 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-07T02:58:52

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

#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!