If log_4 x^2 = 32 what is x.

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freemihai's profile pic

freemihai | College Teacher | (Level 2) Adjunct Educator

Posted on

I will solve the logarithmic equation creating matching bases to the left and to he right of equal.

To the left side tells us what is the base of logarithm, so I will create the logarithm in base 4 to the right.

First, I will consider 32 multiplied by 1. Notice that it does not change a thing this multiplication. Take a look!

log_4 x^2 = 32*1

Now, I will put log_4 (4) in the place of 1:

log_4 x^2 = 32*log_4 (4)

All I need to do is to make use of logarithmic rule of powers and to transform 32 in the exponent of number 4:

log_4 x^2 = log_4 4^32

Now the matching bases make the things much more easy because you just need to set equal the arguments of logarithms. Take a look:

x^2 = 4^32

We must solve for x and not for x^2, so you should consider taking square roots to the left and to the right, same time:

sqrt(x^2) = sqrt(4^32)

|x| = 4^16

Take out the modulus and you will get the final results. Watch out because there are two results, one positive, one negative!

x = +-4^16

The final answer is x = +-4^16.

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation `log_4 x^2 = 32` has to be solved for x.

`log_4 x^2 = 32`

=> `x^2 = 4^32`

=> `x^2 = (2^2)^32`

=> `x^2 = (2^32)^2`

=> `x = +-2^32`

The solution of the equation is `x = +-2^32`

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