Can someone please explain how to solve a log equation with an example.

- anonymous

Can someone please explain how to solve a log equation with an example.

- katieb

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- anonymous

how 'bout u pick an example....

- anonymous

otherwise i'll choose the easiest...:)

- anonymous

OK ...
\[5^{x - 3} . 3^{2x-8} = 225\]

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## More answers

- anonymous

How to solve this using log?
I'm new so I have no idea.

- anonymous

I mean new to log.

- anonymous

pls help.

- lgbasallote

welcome to logarithms!

- anonymous

Yeah thanks!

- anonymous

I guess you can solve this without using Logs..

- anonymous

But I want to know how to do it with log.

- anonymous

Okay then all here will explain you how to do this...

- anonymous

15^(x-3)+(2x-8)=225
now we take ln
(3x-11)15= ln225

- anonymous

What is In?
And how did you get 15^(x-3)+(2x-8)=225?

- anonymous

15*ln(3x-11)=ln 225

- anonymous

by multi 3 into 5 to get 15
and the exponent it's added

- anonymous

But how can you just multiply the base and add the exponents?

- anonymous

Use the following formulas of Logs:
\[\large Log(a \times b) = Log(a) + Log(b)\]
\[\large Log(a)^b = b.Log(a)\]

- anonymous

before we take the logarithem

- anonymous

it's from exponent experepties

- anonymous

Why did you delete that @dpaInc?

- anonymous

because i don't wanna take away from these guys awesome explanations....:)

- anonymous

See, take Log both the sides you will get:
\[\large Log(5^{x-3}.3^{2x-8}) = Log(225)\]
Now use the formulas I have written above..

- anonymous

Use the first formula only...

- anonymous

\[\log(5^{x-3}) . \log(3^{2x-8}) = \log 225 \]

- anonymous

Now Check my first formula carefully...

- anonymous

@pratu043 trust take my answer

- anonymous

like what waterineyes did now

- anonymous

That's what I've done here right @waterineyes?

- anonymous

I give you an example:
\[\large Log(2 \times 3) = Log(2) + Log(3)\]
Go by this procedure...

- anonymous

I've already done that. So now we should use the second one?

- anonymous

You have done it wrongly my friend...

- anonymous

See check the formula..
After formula the result will be in sum form and your result is not in sum form..
Is there any + sign in your result???

- anonymous

Oops.
\[\log(5^{x-3}) + \log(3^{2x-8}) = \log225\]

- anonymous

Yes now you are right...
Now 225 is square of what number???

- anonymous

15

- anonymous

So can you write 225 as \(15^2\) on right hand side.??
If yes then do it..

- anonymous

\[\log(5^{x-3}) + \log(3^{2x-8}) = \log15^{2}\]

- anonymous

Now use the power rule of Logarithms that is the Second Formula that I have given above..
Can you use that??

- anonymous

I give you an example of that too:
\[\large Log(x)^4 = 4.Log(x)\]

- anonymous

\[(x-3)\log5 + (2x-8)\log3 = 2\log15\]

- anonymous

Can you write 15 as 3*5??

- anonymous

Yes.
\[(x-3)\log5 + (2x-8)\log3 = 2\log(5 * 3)\]

- anonymous

Then use it and use the first formula again on right hand side...

- anonymous

\[(x-3)\log5 + (2x-8)\log3 = 2\log5 + 2\log3\]

- anonymous

Now what you have to do is to compare the coefficients of \(Log3\) and \(Log5\) both the sides can you do that??

- anonymous

Tell me the coefficient of Log3 on Left hand side and right hand side...

- anonymous

On LHS its 2x - 8 and on RHS its 2.

- anonymous

Just equate them and find x from it..

- anonymous

Similarly equate the coefficients of log5 and then also find x you will get the same x in both the case...

- anonymous

No no...

- anonymous

\[(2x-8)\log3 = 2\log3\]

- anonymous

I said to equate the coefficients buddy..
Coefficients are 2x - 8 and 2..
you should equate this only...

- anonymous

So you should just do 2x-8 = 2?

- anonymous

x = 5

- anonymous

See, if :
\[\large xlog3 = ylog3\]
implies x = y.
Getting??

- anonymous

Yes!!

- anonymous

Yes you have found x = 5 that is absolutely right...

- anonymous

Thanks a lot!!

- anonymous

Now equate the coefficients of log5..

- anonymous

x - 3 = 2
x = 2 + 3
= 5

- anonymous

Yes got the same value of x so you are right..
If anyhow x values are coming different then either question is wrong or your solution is wrong..
Getting??

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