Solve for $x$
$$ \log_3(3x + 2) = \log_9(4x + 5) $$
I changed the bases of the logs
$$ \frac {\log_{10}(3x + 2)} {\log_{10}(3)} = \frac {\log_{10}(4x + 5)} {\log_{10}(9)} $$
Now I'm stuck, I don't know how to eliminate the logs.
On WolframAlpha I've seen that $\dfrac {\log_{10}(3x + 2)} {\log_{10}(3)} = 0$ gives $ x = -\frac{1}{3}$.
Do you know how does this equation and the equation above can be solved?
Thanks in advance, first question here :D
EDIT: You all solved the first equation, but I still don't understand how to solve the second one (the one solved by WolframAlpha).
$\endgroup$ 26 Answers
$\begingroup$All you need is $\log_9 = \frac12\log_3$: $$ \log_3 (3x+2) = \frac12\log_3(4x+5)\Longrightarrow (3x+2)^2 = 4x+5 $$ Could you proceed?
$\endgroup$ $\begingroup$Notice, formula $$\color{blue}{log_{b^n}(a)=\frac{1}{n}log_{b}(a)}$$ Now, we have $$log_{3}(3x+2)=log_{9}(4x+5)$$ $$\implies log_{3}(3x+2)=log_{3^2}(4x+5)$$ $$\implies log_{3}(3x+2)=\frac{1}{2}log_{3}(4x+5)$$ $$\implies 2 log_{3}(3x+2)=log_{3}(4x+5)$$ $$\implies log_{3}(3x+2)^2=log_{3}(4x+5)$$ $$\implies (3x+2)^2=4x+5$$ $$\implies 9x^2+12x+4=4x+5$$ $$\implies 9x^2+8x-1=0$$ $$\implies (9x-1)(x+1)=0$$ $$\implies x=\frac{1}{9} \ \text{&} \ x=-1$$ Now, substituitng $x=-1$ , we get $LHS=\log(3(-1)+2)=\log(-1)$
But log is not defined for negative number hence the correct solution is $\color{blue}{x=\frac{1}{9}}$
$\endgroup$ 2 $\begingroup$$$\log_3(3x + 2) = \log_9(4x + 5)$$ Convert the RHS to base $3$ to get $$\log_3 (3x+2) = \frac{\log_3 (4x+5)}{\log_3 3^2}$$
So that you get $$2\log_3 (3x+2) = \log_3 (4x+5)$$
The power law for logarithms yields $$\log_3 (3x+2)^2 = \log_3 (4x+5)$$
Now, "cancelling out the logarithms" gives you $$\bbox[10px, border:solid blue 1px]{(3x+2)^2 = 4x+5}$$ which is an easy quadratic to solve.
$\endgroup$ $\begingroup$$$ \log_3(3x + 2) = \log_9(4x + 5) $$
You can use this:
$$\log_b a = \frac{\log_c a}{\log_c b}$$
Change to base 9: $$ \log_3 (3x+2) = \frac{\log_9 (3x+2)}{\log_9 3}$$
$$ \log_9 (3x+2) = \frac{1}{2}\log_9 (4x+5)$$
$$ 2\log_9 (3x+2) = \log_9 (4x+5)$$
$$ \log_9 (3x+2)^2 = \log_9 (4x+5)$$
$$(3x+2)^2 = 4x+5$$
$$ 9x^2 + 8x -1 = 0$$
$$(9x-1)(x+1)=0$$
$$ x = \frac{1}{9} $$
Note: when $x =-1$, $\log_3 (3x+2)= \log_3(-1)$ which has no real solution
$\endgroup$ 2 $\begingroup$For your second question:
$$\Rightarrow\dfrac{\log_{10}{(3x+2)}}{\log_{10}{(3)}}=0$$
$$\Rightarrow\log_{10}{(3x+2)}=0$$
Notice $\color{blue}{10^0=1}$
Hence,
$$\Rightarrow\log_{10}{(3x+2)}=\log_{10}{(1)}$$
Equating logarithm of same bases,
$$\Rightarrow 3x+2=1$$
$$\color{red}{\therefore\space x=\dfrac{-1}{3}}$$
I guess you should red a little more on >>>
As for the first you should know that $\large\log_{b^n}{(a)} = \dfrac{1}{n} \log_b{(a)}$
$\endgroup$ 1 $\begingroup$We have $$\log_3(3x + 2) = \log_9(4x + 5)\implies \frac{\ln(3x+2)}{\ln3}=\frac{\ln(4x+5)}{\ln3^2}$$ and since $\ln3^2=2\ln3$, we get $$2\ln(3x+2)=\ln(4x+5)$$ and taking exponentials gives $$(3x+2)^2=4x+5\implies 9x^2+8x-1=(9x-1)(x+1)=0$$ so $$x=\frac19\quad\text{and}\quad x=-1$$ But $\ln(3\cdot(-1)+2)=\ln(-1)$ is undefined so $x=1/9$ is the only solution.
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