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This is my first post and I honestly just want a second opinion on my answer to a question I got incorrect on an exam before I go arguing over it with my professor.

Basically, is this mathematically correct to do?

$\log y = B_0 + B_1 \log x_1 - 2 B_2 \log x_2 + B_2 \log x_2 + B_3 + 2 B_2 \log x_2$

and reordering it to

$\log y = B_0 + (B_1-2 B_2)\log \frac {x_1}{x_2} + 3 B_2 \log x_2 + B_3$

I know it's a simple question for a website of such caliber, but I've always thought the subtraction of two logarithmic functions can be combined like that

Karolis Juodelė
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user143821
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  • Unfortunately what you wrote is wrong. Example: $3\log(4) - 2\log(2) "=" \log(2)$ using your reasoning. But it's $4\log(2)...$ – user88595 Apr 17 '14 at 16:24

1 Answers1

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The error in your answer was at this part: $B_1(\log x_1)-2B_2(\log x_2)= (B_1-2B_2)\log\left(\dfrac{x_1}{x_2}\right)$

A counterexample:

Consider $\ln e-2\ln 1$.

Then according to what you wrote, it should be $(1-2)\ln\left(\frac{e}{1}\right)=-1\cdot 1=-1$

But, $\ln e-2\ln 1=1-(2\cdot 0)=1-0=1$

Of course $-1\neq 1$.

Hence in general, $a\log x_1-b\log x_2\neq (a-b)\log \left(\frac{x_1}{x_2}\right)$

Sujaan Kunalan
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