# Thread: Some Basic Logarithims Questions!

1. ## Some Basic Logarithims Questions!

Ok guys, just brushing up on some review questions and a few have stumped me. Appreciate it if you can help me out WITH HOW you solved the question. Thanks! Now the questions...

1) $\displaystyle \frac {\log_{11} 64}{\log_{11} 16}$

**This one I thought was really easy but I got it wrong... So, If the base is the same (11), can't I just divide 64/16=4? Or, wait, can this be written as:

$\displaystyle {\log_{11} 64}$- $\displaystyle {\log_{11} 16}$

So $\displaystyle {\log_{11} 48}$ would be my answer?

2) Simplify
$\displaystyle ln(a^2-b^2)-ln(a+b)$

as much as possible.

** This again is a difference question so I can rewrite this as a quotient and solve? Right track?

3) If $\displaystyle x=e^y.$ write $\displaystyle ye^y$ as a simple function of x.

** I suck at word problems or interpreting you can say. Kind of lost on this one but I am thinking, re write this as an exponential function? Not sure, just throwing it out there!

2. Originally Posted by mvho
Ok guys, just brushing up on some review questions and a few have stumped me. Appreciate it if you can help me out WITH HOW you solved the question. Thanks! Now the questions...

1) $\displaystyle \frac {\log_{11} 64}{\log_{11} 16}$

**This one I thought was really easy but I got it wrong... So, If the base is the same (11), can't I just divide 64/16=4? Or, wait, can this be written as:

$\displaystyle {\log_{11} 64}$- $\displaystyle {\log_{11} 16}$

So $\displaystyle {\log_{11} 48}$ would be my answer?
No, neither method is right.
$\displaystyle \frac{\log a}{\log b} \neq \log a - \log b$. You're confusing with
$\displaystyle \log \left(\frac{a}{b}\right) = \log a - \log b$.

\displaystyle \begin{aligned} \frac{\log_{11} 64}{\log_{11} 16} &= \frac{\log_{11} 4^3}{\log_{11} 4^2} \\ &= \frac{3\log_{11} 4}{2\log_{11} 4} \\ &= \frac{3}{2} \end{aligned}

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3. Originally Posted by mvho
2) Simplify
$\displaystyle ln(a^2-b^2)-ln(a+b)$

as much as possible.

** This again is a difference question so I can rewrite this as a quotient and solve? Right track?
Yes:

\displaystyle \begin{aligned} \ln (a^2-b^2)- \ln(a+b) &= \ln \left(\frac{a^2-b^2}{a + b}\right) \\ &= \ln \left(\frac{(a + b)(a - b)}{a + b}\right) \\ &= \ln (a - b) \end{aligned}
This is all assuming, of course, that everything inside the logarithms are positive numbers.

01

4. Originally Posted by mvho

3) If $\displaystyle x=e^y.$ write $\displaystyle ye^y$ as a simple function of x.

$\displaystyle x=e^y \Rightarrow y=ln(x)$

now $\displaystyle ye^y = ln(x)e^{ln(x)} = x\times ln(x)$

5. Wow, Thanks alot guys. I feel silly now seeing how simple it is. For question #1, find the same base and solve.

For question #2, I should've caught the difference of squares. It seems obvious now that you have solved it. LOL