1. ## Solving Logarithmic Equations

I need help trying to figure out how to solve

log (2-3x) + log (3-2x) = 1.5

logbase7(1-x) - logbase7(x+2) = logbase7(x^2)

Thank You!

2. Hello,

log (2-3x) + log (3-2x) = 1.5
You know that log(a)+log(b)=log(ab)

So $\log(2-3x)+\log(3-2x)=\log((2-3x)(3-2x))=1.5$

And you can continue by developping what's in brackets and taking the exponential (beware of the domain of the function)

logbase7(1-x) - logbase7(x+2) = logbase7(x^2)
Since $\log_{x} y = \frac{\log(y)}{\log{x}}$, you can transform the previous equation :

$\log_7 (1-x) - \log_7 (x+2) = log_7 (x^2) \Longleftrightarrow \log (1-x) - \log(x+2) = \log(x^2)$

Since log(a)-log(b)=log(a/b), it's the same as :

$\log(\frac{1-x}{x+2}) = \log(x^2)$

And if log(a)=log(b), then a=b

3. Originally Posted by Moo
Hello,

You know that log(a)+log(b)=log(ab)

So $\log(2-3x)+\log(3-2x)=\log((2-3x)(3-2x))=1.5$

And you can continue by developping what's in brackets and taking the exponential (beware of the domain of the function)

I don't know what to do after that. (log((2-3x)(3-2x))=1.5) am I supposed to subtract 1.5 and get 6x^2- 13x + 4.5? And go from there?

4. Nope,

You have log(....)=1.5

So "...." = exp(1.5)

5. ## Correct?

Hi!

Is this correctly simplified?

log (6-13x+6x^2) = 1.5 then..

6x^2 - 13x + 6 = 10^1.5

then solve this second degree equation..

I get x1 = 3,41 x2= -1,25