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

Mathematics
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how 'bout u pick an example....
otherwise i'll choose the easiest...:)
OK ... \[5^{x - 3} . 3^{2x-8} = 225\]

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Other answers:

How to solve this using log? I'm new so I have no idea.
I mean new to log.
pls help.
welcome to logarithms!
Yeah thanks!
I guess you can solve this without using Logs..
But I want to know how to do it with log.
Okay then all here will explain you how to do this...
15^(x-3)+(2x-8)=225 now we take ln (3x-11)15= ln225
What is In? And how did you get 15^(x-3)+(2x-8)=225?
15*ln(3x-11)=ln 225
by multi 3 into 5 to get 15 and the exponent it's added
But how can you just multiply the base and add the exponents?
Use the following formulas of Logs: \[\large Log(a \times b) = Log(a) + Log(b)\] \[\large Log(a)^b = b.Log(a)\]
before we take the logarithem
it's from exponent experepties
Why did you delete that @dpaInc?
because i don't wanna take away from these guys awesome explanations....:)
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..
Use the first formula only...
\[\log(5^{x-3}) . \log(3^{2x-8}) = \log 225 \]
Now Check my first formula carefully...
@pratu043 trust take my answer
like what waterineyes did now
That's what I've done here right @waterineyes?
I give you an example: \[\large Log(2 \times 3) = Log(2) + Log(3)\] Go by this procedure...
I've already done that. So now we should use the second one?
You have done it wrongly my friend...
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???
Oops. \[\log(5^{x-3}) + \log(3^{2x-8}) = \log225\]
Yes now you are right... Now 225 is square of what number???
15
So can you write 225 as \(15^2\) on right hand side.?? If yes then do it..
\[\log(5^{x-3}) + \log(3^{2x-8}) = \log15^{2}\]
Now use the power rule of Logarithms that is the Second Formula that I have given above.. Can you use that??
I give you an example of that too: \[\large Log(x)^4 = 4.Log(x)\]
\[(x-3)\log5 + (2x-8)\log3 = 2\log15\]
Can you write 15 as 3*5??
Yes. \[(x-3)\log5 + (2x-8)\log3 = 2\log(5 * 3)\]
Then use it and use the first formula again on right hand side...
\[(x-3)\log5 + (2x-8)\log3 = 2\log5 + 2\log3\]
Now what you have to do is to compare the coefficients of \(Log3\) and \(Log5\) both the sides can you do that??
Tell me the coefficient of Log3 on Left hand side and right hand side...
On LHS its 2x - 8 and on RHS its 2.
Just equate them and find x from it..
Similarly equate the coefficients of log5 and then also find x you will get the same x in both the case...
No no...
\[(2x-8)\log3 = 2\log3\]
I said to equate the coefficients buddy.. Coefficients are 2x - 8 and 2.. you should equate this only...
So you should just do 2x-8 = 2?
x = 5
See, if : \[\large xlog3 = ylog3\] implies x = y. Getting??
Yes!!
Yes you have found x = 5 that is absolutely right...
Thanks a lot!!
Now equate the coefficients of log5..
x - 3 = 2 x = 2 + 3 = 5
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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