x^(1+lg3 x) = 9x^2

First we will apply the logarithm for both sides>

==> lg3 x^(1+ lg3 x) = lg 3 ( 9x^2)

==> (1+lg3 x) *lg3 x = lg3 (9x^2)

==> (1 + lg3 x) * lg3 x = lg3 9 + lg3 x^2

==> lg3 x + (lg3 x)^2 = 2 + 2lg3 x

==> lg3 x)^2 - lg3 x - 2 = 0

Let y= lg3 x

==> y^2 - y - 2 = 0

==> ( y-2)(y+1) = 0

==> y1= 2 ==> 2= lg3 x ==> x = 3^2 = 9

==> y2= -1 ==> -1 = lg3 x ==> x = 3^-1 = 1/3

Then the answer is:

**x = { 1/3, 9}**

We have to find x if x^(1+log3 x)=9x^2

x^(1+log3 x)=9x^2

Take the log to the base 3 on both the sides.

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

=> (1 + log3 x)* log3 x = 2 log3 3x

=> (1 + log3 x)* log3 x = 2 (log3 3 + log3 x)

=> (1 + log3 x)* log3 x = 2 (1 + log3 x)

=> (1 + log3 x)(log3 x - 2) =0

So we have log3 x = -1 and log3 x = 2

=> x = 3^-1 and x = 3^2

=> x = 1/3 and x = 9

**Therefore x is 1/3 and 9.**

We'll take logarithms both sides:

log3 [x^(1+log3 x)] = log3 (9x^2)

We'll apply the power rule to the left side:

(1+log3 x)*(log3 x) = log3 (9x^2)

We'll apply the product rule to the right side:

(1+log3 x)*(log3 x) = log3 9 + log3 (x^2)

We'll remove the brackets:

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

We'll move all terms to one side:

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

We'll combine like terms:

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

We'll note log3 x = t

t^2 - t - 2 = 0

We'll apply quadratic formula:

t1 = [1 + sqrt(1 + 8)]/2

t1 = (1 + 3)/2

t1 = 2

t2 = -1

log3 x = t1

log3 x = 2

x = 3^2

x1 = 9

log3 x = t2

log3 x = -1

x2 = 3^-1

x2 = 1/3

**The solutions are: {1/3 ; 9}.**