# Question about a Logarithmic Property

• B
Gold Member
Say I have ##log_5(x)=log_5\left(\frac{2x+3}{2x-3}\right)##

This means that the value of the LHS and RHS are equal. I take this to mean that "5 raised to some exponent is equal to both x and ##\frac{2x+3}{2x-3}##.

I can now write this as ##x=\frac{2x+3}{2x-3}## because since the function is one-to-one, one output corresponds to exactly one input (the input being the argument).

But what if the case was reversed? That is, the bases were different and the arguments were equal? Could I equate the bases to each other? It would be like saying that two different numbers, raised to some exponent, are equal to the same thing.

Gold Member
I guess I'd like to ask a further question as well: Why do the bases have to be equal to use this property?

Say I have ##log_3(x)=log_5\left(\frac{2x+3}{2x-3}\right)## In having them being equal to each other, is saying that their outputs are the equal. And if their outputs are equal, and it's a one-to-one function, then their inputs should be equal as well and I would be able to write ##x=\frac{2x+3}{2x-3}##

mfb
Mentor
and it's a one-to-one function
You have different one-to-one functions on both sides.
If f(x)=2x and g(x)=3x+1, then f(x)=g(y) doesn't mean x=y. It means 2x=3y+1.

You can convert logarithms into each other: ##\log_n(x) = \log_n(m) \log_m(x)##. Or, equivalently: ##\displaystyle \log_n(x)\log(n) = \log_m(x)\log(n)## where the unmarked logarithms on both side need the same (arbitrary) base.

opus
Gold Member
You have different one-to-one functions on both sides.
If f(x)=2x and g(x)=3x+1, then f(x)=g(y) doesn't mean x=y. It means 2x=3y+1.

You can convert logarithms into each other: ##\log_n(x) = \log_n(m) \log_m(x)##. Or, equivalently: ##\displaystyle \log_n(x)\log(n) = \log_m(x)\log(n)## where the unmarked logarithms on both side need the same (arbitrary) base.
So if the functions have different bases, they're different one-to-one functions. And even if they have the same outputs, that doesn't mean that they have the same inputs and vice versa.

If functions do have the same base, that means that they're the same one-to-one function. In this case, having the same output means that they do in fact have the same input.

Is this a true statement?

Gold Member
I may need to retract that last statement. In one problem in my text, we have:

(i) ##3^{log_{3x}(2)}=2##
(ii) ##log_{3x}(2)=log_3(2)##
(iii) ##3x=3##
(iv) ##x=1##

In this case, the text says to equate the bases similarly to equating exponents of like-bases for exponential equations. However in this case, these logs have different bases. One base is 3x and one is 3.

Mark44
Mentor
I may need to retract that last statement. In one problem in my text, we have:

(i) ##3^{log_{3x}(2)}=2##
(ii) ##log_{3x}(2)=log_3(2)##
(iii) ##3x=3##
(iv) ##x=1##

In this case, the text says to equate the bases similarly to equating exponents of like-bases for exponential equations. However in this case, these logs have different bases. One base is 3x and one is 3.
My advice is to simply use the properties of logs and exponents.
From (i), take ##\log_3## of both sides. This yields (ii) ##\log_{3x}(2)=\log_3(2)##.

To get to the next step, the idea is that if ##log_a(n) = \log_b(n)##, the bases a and b have to be equal.

To see this, let ##u = \log_a(n)##. This means that ##n = a^u##
And let ##v = log_b(n)##, which means that ##n = b^v##.

Since ##a^u = b^v##, and we know that u = v, then we have that ##a^u = b^u##, which means that a has to equal b.

opus
Gold Member
So is this just the exponential property that if ##a^u=a^v##, then ##u=v## because the exponential function is one-to-one, and as such, one input corresponds to only one output? It seems like what you did is just convert the logs into exponents and use this property, and this is okay because you can convert from one form to the other. Is this true?

Mark44
Mentor
So is this just the exponential property that if ##a^u=a^v##, then ##u=v## because the exponential function is one-to-one, and as such, one input corresponds to only one output? It seems like what you did is just convert the logs into exponents and use this property, and this is okay because you can convert from one form to the other. Is this true?
Yes.
One thing I didn't mention is that in the implication ##a^u = b^u \Rightarrow a = b##, both a and b have to be positive. If not, we might have something like ##(-2)^2 = 2^2##, but ##-2 \ne 2##. But since a and b are both log bases, there's the usual restriction that they be positive real numbers not equal to 1.

opus
Gold Member
Excellent. Thank you everyone.