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#