Logarithm Question

[SDPY15]

Ok, if you have a logarithm equation like: log_root(3)(9) + log₂₅(5) then how would you simplify it?

So far I have tried separating the equation into two halves and making each part equal to x so as to derive a way to get a common base, but that isn't working because I don't know how to simplify 9=root(3)^x to get a common base with 5=25^x.

I also understand that you can find the answer by taking the inverse of the eq, but i still don't understand what to do...

 

[Deleted member 4993]

Ok, if you have a logarithm equation like: log_root(3)(9) + log₂₅(5) then how would you simplify it?

So far I have tried separating the equation into two halves and making each part equal to x so as to derive a way to get a common base, but that isn't working because I don't know how to simplify 9=root(3)^x to get a common base with 5=25^x.

I also understand that you can find the answer by taking the inverse of the eq, but i still don't understand what to do...

let \(\displaystyle log_{\sqrt{3}}(9) \ = \ x\)

then

(3^1/2)^x = 9 = 3² → 1/2 * x = 2 → x = 4

And

let \(\displaystyle log_{25}(5) \ = \ y\)

then

(5²)^y = 5 = 5¹ = → 2 * y = 1 → y = 1/2

Then

\(\displaystyle log_{\sqrt{3}}(9) \ + \ log_{25}(5) \ = \ x + y \) and continue.....

 

[HallsofIvy]

First, you understand that is not an equation don't you? Second, you don't need to get a "common base". You can simply write each as a number. \(\displaystyle y= log_a(x)\) is the same as \(\displaystyle x= a^y\) so \(\displaystyle y= log_{\sqrt{3}}(9)\) is the same as \(\displaystyle 9= (\sqrt{3})^y\). \(\displaystyle (\sqrt{3})^2= 3\) of course, and squaring again will give 9: \(\displaystyle (\sqrt{3})^4= 9\) so that \(\displaystyle log_{\sqrt{3}}(9)= 4\).

Similarly, \(\displaystyle y= log_{25}(5)\) is the same \(\displaystyle 5= 25^y\). So what is y?

 

[SDPY15]

let \(\displaystyle log_{\sqrt{3}}(9) \ = \ x\)

then

(3^1/2)^x = 9 = 3² → 1/2 * x = 2 → x = 4

And

let \(\displaystyle log_{25}(5) \ = \ y\)

then

(5²)^y = 5 = 5¹ = → 2 * y = 1 → y = 1/2

Then

\(\displaystyle log_{\sqrt{3}}(9) \ + \ log_{25}(5) \ = \ x + y \) and continue.....

Yes, I understand the 3^1/3 now, but my teacher only wants me to simplify the question into a general statement. In other words we don't have to solve it for x and y, we only have to reduce it into its simplest form. This is where I'm always going wrong because when you add the equations you have to have a common base so you can multiply them to form the statement.

 

[pappus]

Ok, if you have a logarithm equation like: log_root(3)(9) + log₂₅(5) then how would you simplify it?

...

Do you know the base-change-formula of logarithms?

\(\displaystyle \displaystyle{\log_b(a)=\frac{\ln(a)}{\ln(b)} = \frac{\log(a)}{\log(b)}}\)

If so

\(\displaystyle \displaystyle{\log_{\sqrt{3}}(9) + \log_{25}(5)=\frac{\ln(9)}{\ln(\sqrt{3})} + \frac{\ln(5)}{\ln(25)} = \frac{2 \ln(3)}{\frac12 \ln(3)} + \frac{1 \cdot \ln(5)}{2 \ln(5)}}\)

Cancel the common factors and calculate the sum.

 

[SDPY15]

Do you know the base-change-formula of logarithms?

\(\displaystyle \displaystyle{\log_b(a)=\frac{\ln(a)}{\ln(b)} = \frac{\log(a)}{\log(b)}}\)

If so

\(\displaystyle \displaystyle{\log_{\sqrt{3}}(9) + \log_{25}(5)=\frac{\ln(9)}{\ln(\sqrt{3})} + \frac{\ln(5)}{\ln(25)} = \frac{2 \ln(3)}{\frac12 \ln(3)} + \frac{1 \cdot \ln(5)}{2 \ln(5)}}\)

Cancel the common factors and calculate the sum.

I don't want to simplify it into common numbers, it has to be simplified into an equation...

 

[DrPhil]

I don't want to simplify it into common numbers, it has to be simplified into an equation...

Since it was never an equation to begin with, you can't simplify it into one.

It is an expression with two terms. One of the terms simplifies to 4, and the other to 1/2. The simplest possible form for the expression is 4 1/2.

 

[lookagain]

One of the terms simplifies to 4, and the other to 1/2. The simplest possible form for the expression is 4 1/2. \(\displaystyle \ \ \ \) or 9/2 or 4.5

.

 

[SDPY15]

Ah, ok. Thanks everyone for your help! (And sorry about saying equation, it's just a common term that my teach is always using. )