Question

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

Answer 1

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`

Answer 2

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`