How to Solve algebraically Log₂x-Log₂(1-2x)= -3?

I'll see what I get:

log₂(x) - log₂(1 - 2x) = -3

First, the difference of two logs is the same as the log of the quotient:

log₂[x / (1 - 2x)] = -3

Now since we have a base 2 log, having both sides be an exponent over a base of 2 will cancel out the log:

x / (1 - 2x) = 2⁻³

Simplifying:

x / (1 - 2x) = 1/8

Cross multiplying:

1 - 2x = 8x

Add 2x to both sides:

1 = 10x

x = 1/10

It's the right answer. If your answer key says otherwise, your answer key is wrong.

To show this, I can plug in x = 0.1 back into the original equation and we should get close to -3 when we work it out:

log₂(x) - log₂(1 - 2x) = -3
log₂(0.1) - log₂(1 - 2*0.1) = -3
log₂(0.1) - log₂(1 - 0.2) = -3
log₂(0.1) - log₂(0.8) = -3

base change:

ln(0.1) / ln(2) - ln(0.8) / ln(2) = -3

subtract the numerators:

[ ln(0.1) - ln(0.8) ] / ln(2) = -3

Now using a calculator, get decimal approximations:

[-2.302585 - (-0.22314355) ] / 0.69314718 = -3
(-2.302585 + -0.22314355) / 0.69314718 = -3
(-2.07944145) / 0.69314718 = -3
-2.99999987 = -3

It's close enough to be confident in the answer, which again is:

x = 1/10 or 0.1

Log₂x-Log₂(1-2x)= -3
log2 ( x /(1-2x) ) = -3
x /(1-2x) = 2^(-3)
x /(1-2x) = 1/8
8x = 1-2x
10 x = 1
x= 1/10

log(2)[x]-log(2)[1-2x]=-3=>
log(2)[x/(1-2x)]=-3=>
x/(1-2x)=2^(-3)=>
8x=1-2x=>
10x=1=>
x=0.1

Quadratic formula not needed.

Log₂(x( - Log₂(1-2x) = -3

Multiply by -1 and rearrange:
Log₂(1-2x) - Log₂(x) = 3

Since log(A) - log(B) = log(A/B)
Log₂((1-2x)/x) = 3

Since Log₂B = A ⇒ 2ᴬ = B:
2³ = (1-2x)/x
8x = 1 - 2x
10x = 1
x = 0.1
__________________________
Check
Log₂(0.1) - Log₂(1-2(0.1) )
= Log₂(0.1/0.8)
= Log₂(1/8)
= -Log₂(8)
= -3
It checks.

Log (x) - Log (1-2x) = -3

subtracting logs is dividing inside the log

Log [(x/(1-2x))] = -3

Since this is log to base 2, we have

x/(1-2x) = 2^(-3)

(1-2x)/x = 2^3 = 8

1 - 2x = 8x

1 = 10x

x = 1/10