# math What is (log3 x)^2=log3 (x^2) + 3 ?

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

You need to move all terms to one side, such that:

(log_3 x)^2 - log_3 x^2 - 3 = 0

You need to come up with the following substitution, such that:

log_3 x = t

t^2 - 2t - 3 = 0

You may use quadratic formula to find the solutions to quadratic equation, such that:

t_(1,2) = (-(-2)+-sqrt((-2)^2 - 4*1*(-3)))/(2*1)

t_(1,2) = (2 +- sqrt(4 + 12))/2

t_(1,2) = (2 +- 4)/2 => t_1 = 3 ; t_2 = -1

You need to substitute back log_3 x = t_(1,2) such that:

log_3 x = 3 => log_3 x = 3*log_3 3 => log_3 x = log_3 3^3 => x = 3^3 => x = 27

log_3 x = -1 => log_3 x = -1*log_3 3 => log_3 x = lo_3 3^(-1) => log_3 x = log_3 (1/3) => x = 1/3

Hence, evaluating the solutions to the given logarithmic equation yields x = 1/3 and x = 27 .

check Approved by eNotes Editorial

## Related Questions

giorgiana1976 | Student

Since we have the term log3 x^2, we'll use the power property of logarithms:log3 x^2 = 2 log3 x.

The given equation (log3 x)^2 = log3 x^2 + 3 converts to

(log3 x)^2 - 2log3 x - 3 = 0

We'll substitute  log3 x = u.

We'll re-write the equation:

u^2 - 2u -3 =0

Factoring, we'll get:

(u-3)(u+1) =0

We'll put each factor as zero:

u-3 = 0

So, u = 3

u + 1 = 0

u =  -1

But log3 x = 3 => x = 3^3 => x = 27

log3 x = -1 => x = 3^-1 => x = 1/3

Since both solutions are positive, we'll accept them.