5 log x - log x^3 = log ( 1-2x) find x

We'll impose constraints of existence of
logarithms:

x>0

1-2x>0

-2x>-1

2x<1

x<1/2

The
interval of admissible values of x is (0 , 1/2).

5 log x -
log x^3 = log ( 1-2x) 

We'll use power property of
logarithms for the first term:

5 log x = log
x^5

log x^5- log x^3 = log (
1-2x) 

We'll add log x^3 both
sides:

log x^5 = log ( 1-2x)  + log
x^3

We'll use the product property
of  logarithms:

log a + log b = log
a*b

We'll put a=( 1-2x)  and
b=x^3

log ( 1-2x)  + log x^3 = log
x^3*(1-2x) 

The equation will
become:

log x^5 = log
x^3*(1-2x) 

Since the bases are matching, we'll use one to
one property:

x^5 = 
x^3*(1-2x) 

We'll divide by
x^3:

x^2 = 1 - 2x

We'll
subtract 1-2x:

x^2 + 2x - 1 =
0

x1 = [4-sqrt(4+4)]/2

x1 =
(4-2sqrt2)/2

x1 =
2-sqrt2

x2=2+sqrt2

Since
neither of x values belong to the interval of admissible values for x, the equation has
no solutions.