How do you solve log_2 x + log_4 x + log_8 x + log_16 x = 25?

2 Answers
Mar 4, 2016

Solution =>x=2^12

Explanation:

log_2x+log_4x+log_8x+log_16x =25
=> 1/log_x2+1/log_x2^2+1/log_x2^3+1/log_x2^4=25
=> 1/log_x2+1/(2log_x2)+1/(3log_x2)+1/(4log_x2)=25
=>1/log_x2(1+1/2+1/3+1/4)=25
=>log_2x(12+6+4+3)/12=25
=>log_2x=25*12/25=12
=>x=2^12

Mar 4, 2016

We can start by converting all of the log terms to the same base using the rule;

log_cb = logb/logc

logx/log2+logx/log4+logx/log8+logx/log16 = 25

Now we can pull logx out of the numerators.

logx(1/log2 + 1/log4 + 1/log8+1/log16) = 25

Notice that all of the denominators are powers of 2, so we can rewrite as;

logx(1/log2 + 1/log(2^2) + 1/log(2^3)+1/log(2^4)) = 25

Using the identity, log(x^y) = y*logx, we can rewrite the denominators as multiples of log2.

logx(1/log2 + 1/(2log2)+1/(3log2)+1/(4log2)) = 25

Now we can pull log2 out of the denominators.

logx/log2(1+1/2+1/3+1/4) = 25

We can rewrite all of the fractions in term of the lowest denominator.

logx/log2(12/12 + 6/12+4/12+3/12)=25

Sum the terms in parenthesis.

logx/log2(25/12) = 25

We can combine all of the constant terms by multiplying both sides by the inverse of the fraction on the left hand side.

logx/log2(25/12)(12/25) = 25(12/25)

Combine the log terms by reversing the process we did in step one and eliminate all of the constant terms that cancel.

log_2x = 12

Using the definition of a logarithm, we write this statement in terms of x.

x=2^12

Solve.

x=4096