Logarithm (Defined x)

Question

This question is hard for me. Solve the equation:

$ \log(5(4x-7))=\log(5(x-2))+1$

Can anybody help me with this question, please?

Thanks.

Answer 1

The idea to apply $\exp(x)$ to both side and you get

$$e^{\log(5(4x-7))}=e^{(\log(5(x-2))+1)}$$

$$20x-35=e^{\log(5(x-2))}.e^{1}$$

$$\frac{20x-35}{e}=5x-10 $$

$$\text{So }\, x=\dfrac{35+10e}{20-5e}.$$

Answer 2

IF the equation is the following:

$\log_5(4x−7)=\log_5(x−2)+1$

$5^{\log_5(4x−7)}=5^{\log_5(x−2)+1} $

$5^{\log_5(4x−7)}=5^{\log_5(x−2)} \cdot 5 $

$4x-7 = 5 (x-2) $

$4x-7 = 5x - 10 $

$- x = -3 $

$x=3$