solve log base 3 (x^2) - log base 3 (x) =2 what is the value of x?

Question

can someone please tell me how to solve this question. thanks

Answer 1

I'm not sure if you mean [ log[base 3](x) ]^2 or log[base 3](x^2). There's a big difference between the two. I'll assume you mean the first.

[ log[base 3](x) ]^2 - log[base 3](x) = 2

Move the 2 to the left hand side.

[ log[base 3](x) ]^2 - log[base 3](x) - 2 = 0

Factor this like you would factor y^2 - y - 2 (which of course factors as (y - 2)(y + 1) ).

( log[base 3](x) - 2 ) (log[base 3](x) + 1) = 0

Now equate each factor to 0 and solve each as a separate equation.

1) log[base 3](x) - 2 = 0
2) log[base 3](x) + 1 = 0

Let's solve these individually.

1) log[base 3](x) - 2 = 0

log[base 3](x) = 2
Convert to exponential form.

x = 3^2
x = 9

2) log[base 3](x) + 1 = 0
log[base 3](x) = -1
Therefore,
x = 3^(-1), so
x = 1/3

That means our solutions are

x = {9, 1/3}

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If you mean the second, i.e.
log[base 3](x^2) - log[base 3](x) = 2

Then the solution is much simpler. Use the log identity to move the 2 inside of the log outside of the log.

2 log[base 3](x) - log[base 3](x) = 2

Combine like terms.

log[base 3](x) = 2

Which means

x = 3^2
x = 9

Answer 2

Take log to mean log base 3 in the following:-
log x² - log x = 2
log (x² / x) = 2
log x = 2
x = 3²
x = 9

Answer 3

log base 3 (x^2) - log base 3 (x) =

log base 3 (x^2)/(x) =

log base 3 (x) = 2

x = 9