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I have this equation (identity?):

$$3^{3x}=3^{2y+1}$$

I understand that this simplifies to:

$$3x=2y+1$$

So, the base (of the exponentials) cancel out. I want to know, what the process/function of this cancelling out is?

I know that you can cancel out exponents by $n$th-rooting the exponents. So by what function do you cancel out the base?

Please tell me if I should give more information. This is probably a very basic step in dealing with algebraic formulas, but I can't find the specific steps anywhere. I want to know the function so I know what steps to use and how/why they work.

Thanks in advance!

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up vote 9 down vote accepted

You use the logarithm of the given base. For instance, if you have (in the natural base)

$$e^{f(x)}$$ you can take the natural logarithm of that and get

$$\ln\left(e^{f(x)}\right)=f(x)\ln(e)=f(x)$$

So in your equation, you take the $\log_3$ on both sides to obtain $$\log_3\left(3^{3x}\right)=\log_3\left(3^{2y+1}\right) \leftrightarrow 3x\log_3(3)=(2y+1)\log_3(3) \leftrightarrow$$

$$3x=2y+1$$

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Thanks for the reply. I've been reading about the log function here. Am I correct in assuming that ln(x) can be treated as any other term? i.e. It can be canceled out like any other term. – Peter S Feb 28 at 17:51
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@PeterS Yes that's exactly right. If you want to know more about rules for handling logarithms, see here: chilimath.com/algebra/advanced/log/images/… – Lovsovs Feb 28 at 17:52
    
Thanks for the information. I appreciate it. I'm still covering a lot of relatively basic mathematics to prepare for exams in a few months. – Peter S Feb 28 at 17:53
    
@PeterS Good luck! Since you're new to the site: To show your appreciation, you can chose one answer (it of course doesn't have to be mine) to your question that you felt answered it satisfactorily and "accept" it, thereby making the answerer happy as a baby - you do this by clicking the tick under the up/down-vote buttons. – Lovsovs Feb 28 at 17:56
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Second to last equation: log3(x) should be log3(3)? – immibis Feb 29 at 0:02

The function $f(t)=3^t$ is increasing. So if $f(a)=f(b)$ we have $a=b$.

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1  
More generally, you could say exponentials are injective, so this reasoning can be applied to bases $0<b<1$. – MathematicsStudent1122 Feb 29 at 7:27

see for basic understanding i can say you that if bases are same and two numbers are in equality then there powers must be equal

$$2^z=2^{33}$$ this means $$z=33$$ but if it is given like $$2^g=15.(2)^{33}$$ then you cannot write $g=33$

for proving this we have to use logarithms.

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Thanks. I guess for relatively simple problems I should just accept that like bases can cancel out. But if I want to prove it, I have to use logarithms. – Peter S Feb 28 at 17:57
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yes! you are correct – DeNiSkA Feb 28 at 17:57

I feel that you're muddling statements with expressions. Expressions (like $3^{3 \cdot x}$) simplify in a largely mechanical way, whereas statements don't. The statement $3^{3 \cdot x} = 3^{2 \cdot y + 1}$ is equivalent to the statement $3 \cdot x = 2 \cdot y + 1$, and that is proven as follows.

First, we prove that $3^{3 \cdot x} = 3^{2 \cdot y + 1}$ implies that $3 \cdot x = 2 \cdot y + 1$. That comes from the fact that the function $z \mapsto 3^z$ is injective, which requires a lot more machinery to actually prove. From considering the graph of the function, though, it will hopefully be intuitive (hint: it's a strictly increasing function).

Second, we prove that $3 \cdot x = 2 \cdot y + 1$ implies that $3^{3 \cdot x} = 3^{2 \cdot y + 1}$. This is easy, because it follows from the fact that functions map equal values to equal results.

Biïmplication gives the equivalence. It's possible that you only needed one of the two implications (probably the first), but I wanted to make sure.

An alternative to the first part of the proof is to use the $\log_3$ function, as others suggested. We have that $\log_3 3^z = z$ (we could, indeed, have said that $\log_3 3^z$ simplifies to $z$), so applying it to the terms on both sides of the original equation yields two equal terms, specifically $3 \cdot x$ and $2 \cdot y + 1$.

Note that there are plenty of places where there is no direct notion of cancelling out. An example is that the function $x \mapsto x^3-x$ (plot), where it's quite difficult to find anything useful by inverting it.

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