Logarithm equations

[tronez]

Hi

I'm having a hard time getting my head around some logarithm problems I'm trying to solve.

So for example the equation where I have to evaluate x

log₃9x+log₃x=4

How would I go about working this out. I know its probably simple and I have to use to laws but I can't seem to do it.

Thanks

 

[srmichael]

Hi

I'm having a hard time getting my head around some logarithm problems I'm trying to solve.

So for example the equation where I have to evaluate x

log₃9x+log₃x=4

How would I go about working this out. I know its probably simple and I have to use to laws but I can't seem to do it.

Thanks

Show us what you have tried so far so we can see where you messed up.

Thanks!

 

[JeffM]

Hi

I'm having a hard time getting my head around some logarithm problems I'm trying to solve.

So for example the equation where I have to evaluate x

log₃9x+log₃x=4

How would I go about working this out. I know its probably simple and I have to use to laws but I can't seem to do it.

Thanks

As you say, you use the laws. There are several ways to use them here, but I would start here:

\(\displaystyle log_3(9x) + log_3(x) = 4 \implies log_3(9) + log_3(x) + log_3(x) = 4.\) What next?

 

[tronez]

Ok thanks for your reply. So I'm thinking I refer to the law which states the Log A + Log B = Log AB?

So considering I know the AB value I work from here?

Thanks

 

[JeffM]

Ok thanks for your reply. So I'm thinking I refer to the law which states the Log A + Log B = Log AB?

So considering I know the AB value I work from here?

Thanks

Yes as a first step I am suggesting that you use the law \(\displaystyle log_c(A * B) \equiv \log_c(A) + log_c(B).\)

But the reason that I chose that as a first step is something that you leanred almost the first day of algebra: to solve an equation for x, you get all the terms involving x on one side of the equation and all the terms not involving x on the other side of the equation. Do you remember that basic rule?

So I need to separate the 9 and the x. Separating 9 and x was why I used \(\displaystyle log_3(9x) = log_3(9) + log_3(x).\)

But my question to you is: what next from \(\displaystyle log_3(9) + log_3(x) + log_3(x) = 4?\)

 

[tronez]

Yes as a first step I am suggesting that you use the law \(\displaystyle log_c(A * B) \equiv \log_c(A) + log_c(B).\)

But the reason that I chose that as a first step is something that you leanred almost the first day of algebra: to solve an equation for x, you get all the terms involving x on one side of the equation and all the terms not involving x on the other side of the equation. Do you remember that basic rule?

So I need to separate the 9 and the x. Separating 9 and x was why I used \(\displaystyle log_3(9x) = log_3(9) + log_3(x).\)

But my question to you is: what next from \(\displaystyle log_3(9) + log_3(x) + log_3(x) = 4?\)

I haven't looked at this for a day or 2. Ok so now its expanded like this can we just work out the log3(9) which is 3 and considering the answer is 4 the others have to make up whats left?

 

[pka]

I haven't looked at this for a day or 2. Ok so now its expanded like this can we just work out the log3(9) which is 3 and considering the answer is 4 the others have to make up whats left?

No. \(\displaystyle \log_3(9)=2\) .

 

[JeffM]

I haven't looked at this for a day or 2. Ok so now its expanded like this can we just work out the log3(9) which is 3 and considering the answer is 4 the others have to make up whats left?

Well, no \(\displaystyle log_3(9) = log_3(3^2) = 2 \ne 3.\)

Here is the basic law of logs, the definition if you will: \(\displaystyle log_a(b) = c \iff a^c = b.\)

It follows from that definition that \(\displaystyle log_a\left(a^n\right) = n.\)

So \(\displaystyle log_3(9) + log_3(x) + log_3(x) = 4 \implies 2 + 2 * log_3(x) = 4.\) Now what?

 

[tronez]

OK yeah my bad so that would make X = 3? Considering log3(3) =1.

 

[JeffM]

OK yeah my bad so that would make X = 3? Considering log3(3) =1.

Yes \(\displaystyle 2 + 2 * log_3(x) = 4 \implies 2 * log_3(x) = 4 - 2 = 2 \implies log_3(x) = 1 \implies x = 3.\)

Now CHECK.

\(\displaystyle log_3(9 * 3) + log_3(3) = log_3(27) + log_3(3) = log_3(27 * 3) = log_3(81) = log_3\left(3^4\right) = 4.\)