anonymous
  • anonymous
Can someone please explain how to solve a log equation with an example.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
how 'bout u pick an example....
anonymous
  • anonymous
otherwise i'll choose the easiest...:)
anonymous
  • anonymous
OK ... \[5^{x - 3} . 3^{2x-8} = 225\]

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