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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).

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    $\begingroup$ Hint: $$log_3(3x+2)=log_9((3x+2)^2), x>-\frac{2}{3}$$ $\endgroup$ – callculus Jul 22 '15 at 14:13
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$$\log_3(3x + 2) = \log_9(4x + 5)$$ 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.

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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?

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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 $\color{blue}{x=\frac{1}{9}}$

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  • $\begingroup$ $x=-1$ is not a solution, because x has to be greater than $-\frac{2}{3}$ $\endgroup$ – callculus Jul 22 '15 at 14:26
  • $\begingroup$ Thanks, you are right. I omitted further steps. I made correction. $\endgroup$ – Harish Chandra Rajpoot Jul 22 '15 at 14:34
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$$ \log_3(3x + 2) = \log_9(4x + 5) $$

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

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  • $\begingroup$ $x=-1$ is not a solution, because x has to be greather than $-\frac{2}{3}$ $\endgroup$ – callculus Jul 22 '15 at 14:34
  • $\begingroup$ Yes, thanks for your comment, added to solution. $\endgroup$ – John_dydx Jul 22 '15 at 14:36
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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 >>> https://en.wikipedia.org/wiki/Logarithm


As for the first you should know that $\large\log_{b^n}{(a)} = \dfrac{1}{n} \log_b{(a)}$

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  • $\begingroup$ thank you, I didn't see that the denominator could be simplified like in a normal equation. The formula unfortunately isn't in the wiki page that I've read (it's not the english page). $\endgroup$ – AR89 Jul 22 '15 at 17:25
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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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