What is the solution of the equation log15 [1/(3^x + x - 5)]=x*(log15 5 - 1) ?

We'll re-write the left term of the
equation:

 log15 [1/(3^x + x - 5)] =  log15 [(3^x + x -
5)^-1]

We'll apply the power rule of
logarithms:

log15 [(3^x + x - 5)^-1] = -log15 (3^x + x -
5)

We'll remove the brackets from the right
side:

x*(log15 5 - 1) = xlog15 5 -
x

x*(log15 5 - 1) = log15 (5^x) - x*log15
15

x*(log15 5 - 1) = log15 (5^x) - log15
15^x

Since the bases are matching, we'll apply the quotient
rule:

log15 (5^x) - log15 15^x = log15
(5^x/15^x)

log15 (5^x/15^x) = log15
(5^x/5^x*3^x)

We'll simplify and we'll
get;

log15 (5^x/5^x*3^x) = log15
(1/3^x)

log15 (1/3^x) =-log15
3^x

We'll re-write the
equation:

 -log15 (3^x + x - 5) = -log15
3^x

We'll add log15 3^x:

log15
3^x -log15 (3^x + x - 5) = 0

Since the bases are matching,
we'll apply the quotient rule:

log15[3^x/(3^x + x - 5)] =
0

[3^x/(3^x + x - 5)] =
15^0

[3^x/(3^x + x - 5)] =
1

3^x = 3^x + x - 5

We'll
eliminate like terms:

0 = x -
5

x = 5

The
solution of the equation is x = 5.