# Logarithmic equation. Need to know if i am teaching right

Two of my friends is studying for a test. They asked me about a simple question. But they told me that i was wrong on a question. I could be wrong. But i need you guys to make sure that they learn the right stuff. So if i was right. I then can tell them how to do the equations

The question: Assume that $\log x = 3$ and $\log y = 4$ Calculate the following equation $\log x^4 + 2\log y - \log(xy)$

I got it to be

$$\log \left( \frac{x^4 y^2}{x y} \right)$$

Then they just change the $x$ and $y$ to the assumed value. So am i right or am i wrong

• But you don't know what $x$ and $y$ are - the point is to write everything in terms of $\log{x}$ and $\log{y}$. – mdp May 17 '13 at 14:23
• You need to put it into an equation only using $\log x$ and $\log y$. So not using, for example, $\log xy$. – user1729 May 17 '13 at 14:23
• I love the intro. People always complain about people asking questions without giving context. I'm not sure anymore, if I really want the context. :P – Nikolaj-K May 17 '13 at 14:26

But they aren’t given the values of $x$ and $y$: they’re given the values of $\lg x$ and $\lg y$. Specifically, $\lg x=3$ and $\lg y=4$, so

$$\lg x^4+2\lg y-\lg(xy)=4\lg x+2\lg y-(\lg x+\lg y)=12+8-7=13\;.$$

It’s perfectly true that

$$\lg x^4+2\lg y-\lg(xy)=\lg\frac{x^4y^2}{xy}\;,$$

but this does not really help them to solve the problem.

• Unless they figure out our well-kept secret of how to find $x$ from $\log x$ ;-) – fgp May 17 '13 at 14:53
• @fgp: Since the base of the logs isn’t given, that involves considerably more work than it’s worth! – Brian M. Scott May 17 '13 at 14:56

You're right. For logarithms in general it holds that $log(a)+log(b)=log(ab)$ and $log(a)-log(b)=log(a/b)$. Furthermore it holds that $alog(b)=log(b^a)$. Hence, your answer is correct.

• Jemil has correctly combined the expression into a single log, but that does not help in answering the actual question. – Brian M. Scott May 17 '13 at 14:26
• The working is correct, but it doesn't answer the question. – user1729 May 17 '13 at 14:26
• @BrianM.Scott I assumed that Jemil only wanted to know whether his working was correct. Indeed you are right though that it does not help in answering the question. – dreamer May 17 '13 at 14:28