Logarithm (Defined the x)

Question

Can somebody help.I can't get the answer for this question.

Solve the equation $\log_3(2x-1)+\log_2(4)=5$ Thanks for your help!

Answer 1

\begin{align}\log_3(2x-1)+\log_24=5&\iff \begin{cases} x>\frac12\\ \log_3(2x-1)=5-\log_24\end{cases}\\&\iff \begin{cases} x>\frac12\\ 2x-1=3^{5-\log_24}\end{cases}\\&\iff x=\frac12+\frac{3^{5-\log_24}}2\end{align}

Answer 2

Starting with $\log_3(2x - 1) + \log_2(4) = 5$

Note: $\log_2(4) = \log_2(2^2) = 2*\log_2(2) = 2$.

Therefore,

$\log_3(2x-1) = 3$

Note: $a^{\log_a(b)} = b$.

Hence,

$2x - 1 = 3^3$ or $x = 14$

Answer 3

First you have to check the domain! So, we have this condition: $$2x-1>0 \iff x > \frac{1}{2}.$$

Now we will rewrite your equation: $$\log_3(2x-1)+ \log_2(2^2)=5 \iff \log_3(2x-1)+ 2\log_2(2)=5 \iff \log_3(2x-1)+ 2=5$$

$$\\$$

$$\iff \log_3(2x-1)=3 \iff 2x-1 = 3^3 \iff 2x = 28 \iff x=14.$$

Because our solution satisfy the above condition for the domain, we will accept it!