# `log_(1/2)(log_4*(log_5(x^2+9)))=1`  1/2,4 and 5 are in the base

lemjay | High School Teacher | (Level 3) Senior Educator

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`log_(1/2)(log_4(log_5(x^2+9)))=1`

To solve for x, we need to eliminate the outermost logarithm.To do so, let's use the rule `b^(log_b a) = a` .

`log_4(log_5(x^2+9))=1/2`

Apply again the same rule of logarithm to cancel `log_4` .

`4^(log_4(log_5(x^2+9))) = 4^(1/2)`

`log_5(x^2+9) = (2^2)^(1/2)`

`log_5(x^2+9) = 2`

Again, use the same rule above.

`5^(log_5 (x^2+9)) = 5^2`

`x^2+9 = 25`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2 - 16 = 0`

`(x-4)(x+4)=0`

Set each factor to zero and solve for x.

`x-4=0`                   and                 `x+4=0`

`x=4`                                               `x=-4`

Then, substitute values of x to the original equation to verify.

`x=4` ,   `log_(1/2)(log_4(log_5(4^2+9)))=1`

`log_(1/2)(log_4(log_5 25))=1`

`log_(1/2)(log_4 2)=1`

`log_(1/2) (1/2) = 1`

`1=1`    `(True)`

`x=-4` ,   `log_(1/2)(log_4(log_5((-4)^2+9)))=1`

`log_(1/2)(log_4(log_5 25))=1`

`log_(1/2)(log_4 2)=1`

`log_(1/2) (1/2) = 1`

`1=1`  `(True)`

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Hence. `x= 4`  and `x=-4` are the solution to the equation `log_(1/2)(log_4(log_5((-4)^2+9)))=1.`

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