# Thread: 3 problems with logarithms

1. ## 3 problems with logarithms

The following problems are giving me a little grief, mostly because I'm not exactly sure how the solutions should look. Any help is appreciated, I think my biggest problems is not understanding the language of mathematics!

Problem 1 Solved

If $\displaystyle \log_{10}2=a$ and $\displaystyle \log_{10}3=b$, find the exact solution to the equation $\displaystyle 12^{x+2}=18^{x-3}$ in terms of $\displaystyle a$ and $\displaystyle b$.

Problem 2 Solved

Let $\displaystyle r=\log_{b}\frac{8}{45}$ and $\displaystyle s=\log_{b}\frac{135}{4}$. Find an ordered pair of integers $\displaystyle (m, n)$ such that $\displaystyle \log_{b}\frac{32}{5}=mr+ns$.

Problem 3

Given $\displaystyle \log_{18}6=a$, find $\displaystyle log_{18}16$ in terms of $\displaystyle a$.

2. This will give you a start.
$\displaystyle \begin{gathered} 12^{x + 2} = \left( {2^2 } \right)^{x + 2} \left( 3 \right)^{x + 2} = \left( {2^{2x + 4} } \right)\left( {3^{x + 2} } \right) \hfill \\ \log \left( {12^{x + 2} } \right) = \left( {2x + 4} \right)\log (2) + \left( {x + 2} \right)\log (3) \hfill \\ \end{gathered}$

3. Wow that's a huge help!

thanks a lot I'm now doing the same thing to the other side of the equation I'll post back with my answer!

4. I got this far on my answer:

$\displaystyle a(2x+4)+b(x+2)=a(x-3)+b(2x-6)$

But I'm not sure how I could isolate a and b...

I got there just by replace log2 and log3 with a and b on what Plato did, then did the same method that he did on the second expression, $\displaystyle 18^{x-3}$, and again replaced log2 and log3 with a and b and came up with the equality above.

5. So does anyone know how I can finish the first problem (or begin the second 2)?

Did I set up the first one wrong, or do I just not know how to isolate them?

6. Continue the first.
$\displaystyle \begin{gathered} 2ax + 4a + bx + 2b = ax - 3a + 2bx - 6b \hfill \\ ax - bx = - 7a - 8b \hfill \\ x = \frac{{ - 7a - 8b}} {{a - b}} \hfill \\ \end{gathered}$

7. ugh....

I was looking at it all wrong I don't even know why I didn't do that, anyway thanks

8. I wonder if I'm beginning Problem 2 completely wrong

I'm not exactly sure how to set it up

9. $\displaystyle \begin{gathered} r = 3\log (2) - 2\log (3) - \log (5) \hfill \\ s = \log (5) + 3\log (3) - 2\log (2) \hfill \\ \hfill \\ 3r = 9\log (2) - 6\log (3) - 3\log (5) \hfill \\ 2s = 2\log (5) + 6\log (3) - 4\log (2) \hfill \\ 3r + 2s = 5\log (2) - \log (5) = \log \left( {\frac{{32}} {5}} \right) \hfill \\ \end{gathered}$

10. Does anyone have any kind of hint for me of what I have to do to solve the third problem?

11. Originally Posted by mikedwd

Problem 3

Given $\displaystyle \log_{18}6=a$, find $\displaystyle log_{18}16$ in terms of $\displaystyle a$.
Having $\displaystyle \log _{18}6=a,$ it's not hard to prove that $\displaystyle \log _{2}3=\frac{1-a}{2a-1},$ hence

$\displaystyle \log _{18}16=\frac{4}{\log _{2}18}=\frac{4}{1+2\cdot \dfrac{1-a}{2a-1}}=8a-4,$

as required.

12. Thanks a lot! It seems so clear now that I see it, but this should help me a lot with more problems to come.

I have one more question, though.

On this problem:

Find all real numbers $\displaystyle x$ for which $\displaystyle \frac{3^{\sqrt{12x}}+3}{4}=3^{\sqrt{3x}}$.

I did as follows:

$\displaystyle 3^{\sqrt{12x}}=4*3^{\sqrt{3x}}-3$

$\displaystyle \frac{3^{\sqrt{12x}}}{3^{\sqrt{3x}}}=1$

$\displaystyle 3^{\sqrt{12x}-\sqrt{3x}}=1$

$\displaystyle 3^{\sqrt{9x}}=1$

$\displaystyle x=0$

Is this the only solution or are there others? This was one of two questions were it asked for "all real numbers", and the other one had multiple solutions.

I think that 0 is the only solution, but how can I be sure?