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

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### 3 Answers

(log3 x)^2=log3 (x^2) + 3

First we know that log a^b = b*log a

==> (log3 x)62 = 2log3 x + 3

Now, we will move all terms to the left side so the right side is 0:

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

Now let us assume that:

log3 x = y

==? y^2 - 2y - 3 = 0

Now we will factor:

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

==> y1= 3 ==> log3 x = 3 ==> x = 3^3 = 27

==> y2= -1 ==> log3 x = -1 ==> x = 3^-1 = 1/3

Then the answer is:

**x = { 1/3 , 27}**

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.**

To solve (log 3 x)^2 = log3 (x^2) +3

We know that log 3 (x^2) = 2log3 x by law of logarithms loa^m = m loga.

Therefore the given equation becomes:

(log3 x)^2 = 2log3 x +3.

t^2 = 2t +3 , where t = log3 x.

t^2-2t-3 = 0.

(t+1)(t-3) = 0.

t+1 = 0 , or t-3 = 0.

t = -1 a or t = 3.

t= 3 gives log 3 x= 3. Or x = 3^3 =27.

t =-1 does give any real solution.

Therefore x = 27 is the solution.