# Thread: I'm having some trouble with a logarithmic equation

1. ## I'm having some trouble with a logarithmic equation

I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

Solve for x in the logarithmic equation:
log7 - log(4x + 5) + log(2x - 3) = 0
Give exact answers and be sure to check for extraneous solutions.
The key says x = 13/5

2. Originally Posted by Dirtsa
I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

Solve for x in the logarithmic equation:
log7 - log(4x + 5) + log(2x - 3) = 0
Give exact answers and be sure to check for extraneous solutions.
The key says x = 13/5
$\displaystyle \log{7} - \log(4x + 5) + \log(2x - 3) = 0$

$\displaystyle \log\left[\frac{7(2x-3)}{4x+5}\right] = 0$

$\displaystyle \frac{7(2x-3)}{4x+5} = 1$

$\displaystyle 7(2x-3) = 4x+5$

finish

3. Originally Posted by Dirtsa
I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

Solve for x in the logarithmic equation:
log7 - log(4x + 5) + log(2x - 3) = 0
Give exact answers and be sure to check for extraneous solutions.
The key says x = 13/5
Use the log-rules

1. $\displaystyle \log(a)-\log(b)=\log\left(\frac ab\right)$ and

2. $\displaystyle \log(a)+\log(b)=\log\left( a\cdot b\right)$

3. $\displaystyle \log7 - \log(4x + 5) + \log(2x - 3) = 0~\implies~ \log\left(\dfrac7{4x+5} \cdot (2x-3)\right)=0$

Now use the logs (the RHS of the equation is a log too) with the base 10:

$\displaystyle \dfrac7{4x+5} \cdot (2x-3) = 1~\implies~14x-21 = 4x+5$

Solve for x.

4. ## thank you

thanks for the help