Question

Hii, i was absent today in my math class and I missed the whole lesson teacher signed us a classwork but I am totally confused please shows me how to start this question 18 I need help.

Answer 1

Given the logarithmic equation below

$\log (3x-2)+\log (x-1)=\log 2x$

We can find the value of x below.

Explanation

This can be done using the addition law of logarithm

$\text{Log a + log b=log(axb)}$

Therefore;

$\begin{gathered} \log (3x-2)+\log (x-1)=\log 2x \\ \log (3x-2)(x-1)=\log 2x \\ \log (3x^2-3x-2x+2_{})=\log 2x \\ \sin ce\text{ log a =log b }\rightarrow a=b \\ \therefore3x^2-5x+2=2x \end{gathered}$

The next step is to solve the resulting quadratic equation.

$\begin{gathered} 3x^2-5x+2=2x \\ 3x^2-5x-2x+2=0 \\ 3x^2-7x+2=0 \\ By\text{ factorization we have;} \\ 3x^2-6x-x+2=0 \\ 3x(x-2)-1(x-2)=0 \\ (3x-1)(x-2)=0 \\ \therefore x=\frac{1}{3};x=2 \end{gathered}$

The values of x are

Answer

$x=\frac{1}{3};x=2$