# Proving a log by induction

I had a test recently and there was a log question that was $$3\log_3(x) - 4\log_3(x) + 1/2\log_3(x).$$ When I solved it I got $$\log_3 \left(\frac{1}{\sqrt{x}}\right).$$ My teacher says that is incorrect, so I asked many people and they said I was correct. She said the only way I could get points back was if I prove it by induction, which I do not know what that is. I was wondering if my answer was right and if so how would I prove it by induction if thats possible? Thanks for all your help. Its critical I get these points back

• It would be good if you could show your work. What answer did the teacher give? – user84413 Sep 15 '15 at 17:07
• Either you are kidding us or the people who told you about induction were kidding you. – Aretino Sep 15 '15 at 17:14
• And more over, what was $exactly$ the question? From my understanding it looks like "condense into one logarithm" for which no induction is needed. – imranfat Sep 15 '15 at 17:15
• Your teacher's wrong. Your answer is right and induction has nothing to do with it. If the teacher won't budge then complain to someone with more authority. – Qiaochu Yuan Sep 15 '15 at 17:19
• I would guess your teacher has the answer as $-\frac{1}{2}\log_3(x)$ and didn't realize it is the same answer.... – N. S. Sep 15 '15 at 17:22

$$3\log_3(x) - 4\log_3(x) + \frac 1 2 \log_3(x)$$

What is written above does not express any math problem, but if you had said that some words above it say "Write this as a single logarithm.", then you'd have a math problem to work on. (One of the flaws (or worse than flaws) of our system of coercing masses of people to learn topics in mathematics whose reason for inclusion in the curriculum is only that they're used in later subjects that most students never take is that they have these strange distorted ways of thinking (if "thinking" it may be called) in which students think that what is written above expresses a problem even when no words accompany it.)

Now notice that it says $$3 L - 4L + \frac 1 2 L.$$ You can do that because the three $\text{“}L\text{''}$s are all the same: all three are $\log_3 x$.

Next, the distributive law is $$\left( 3 - 4 + \frac 1 2\right) L.$$ Arithmetic tells us that this is $$\frac{-1}2 L.$$ All of that is done without knowing anything about logarithms. But the next step requires some knowledge of logarithms: $$\frac{-1} 2 \log_3 x = \log_3 (x^{-1/2}).$$ If it had said "Write this expression as a single logarithm", then we're done with that.

Knowledge of exponents tells us that $x^{-1/2} = \dfrac 1 {\sqrt x}$. Hence the expression we're trying to write a just one logarithm is $$\log_3 \frac 1 {\sqrt x}.$$

Induction is not an appropriate technique for any of this.

• thank you for this. My teacher is not very smart. I think she just said that because we do not know what that is. Thanks for this – Ryan Sep 15 '15 at 17:46
• By showing this to her, would I prove that my answer is correct? – Ryan Sep 15 '15 at 17:52
• @Ryan : Perhaps you would. One qualm I have is the possibility that you misunderstood what your teacher said and consequently we don't really have that story. But your other theory --- that she tried to pull the wool over your eyes, is also something that happens. ${}\qquad{}$ – Michael Hardy Sep 15 '15 at 19:30
• @Micheal Hardy I can tell you that I did understand what she said. She did the problem a little different. She did the addition part of the problem first and then applied that to the subtraction. Making it log3(1/x^1.5). I know that is incorrect and the only way it could be correct is if there were parentheses which there was not – Ryan Sep 15 '15 at 23:46
• @MichealHardy Because induction does not apply, how could I prove that log a - log b + log c = log ac/b. – Ryan Sep 16 '15 at 3:03

Shifting the powers, you get:

$$\log_3(x)^3 - \log_3(x)^4 + \log_3(x)^.5$$ $$= \log_3( x^3 / x^4) + \log_3(x)^.5$$ $$= \log_3 \frac 1 x + \log_3(x)^{0.5}$$ $$= \log_3 \frac 1 x \cdot x^{0.5}$$ $$= \log_3 \frac 1 {x^{0.5}} == \text{CORRECT!}$$

Edit: Saw wrong question. OP answer was correct after all

• I don't think one should write $=\text{CORRECT}$. The expression preceding $\text{“}=\text{''}$ is not equal to something called "CORRECT". ${}\qquad{}$ – Michael Hardy Sep 15 '15 at 17:28
• @MichaelHardy how about == ? – xx53xx Sep 15 '15 at 17:32
• This is what I had but she says its incorrect. How can i prove it is correct? She said I have to prove it through induction. I am unsure what that is – Ryan Sep 15 '15 at 17:47