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

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

First let us rewrtie:

we know that:

a log b = log b^2

==> 5log x= log x^5

==> log x^5 - log x^3 = log 1-2x

Also, we know that:

log a - log b = log a/b

==> log x5/x^3 = log (1-2x)

==> log x^2 = log (1-2x)

But, if log a = log b ==> a= b

==> x^2 = 1- 2x

==> x^2 + 2x - 1 = 0

=> (x-1)^2 = 0

==> x= 1

But log (1-2x) is not defined when x= 1

==> There are no solutions for the equation.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

5logx - logx^3 = log (1-2x).

We use    the properties of logarithms: logx^m = m logx and  loga-logb = loga/b.

Therefore we  replace ogx^3 by 3logx in the given equation.

5logx - 3logx = log(1-2x)

2logx = log(1-2x).

logx^2 = log(1-2x)

We take antilog:

x^2 = 1-2x

x^2+2x-1 = 0

x = {-2 +or -sqrt(2^2-4*1*(-1))}/2

x = (-1 +sqrt2} or x = (-1-sqrt2).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.