pratu043
Can someone please explain how to solve a log equation with an example.
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dpaInc
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how 'bout u pick an example....
dpaInc
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otherwise i'll choose the easiest...:)
pratu043
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OK ...
\[5^{x - 3} . 3^{2x-8} = 225\]
pratu043
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How to solve this using log?
I'm new so I have no idea.
pratu043
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I mean new to log.
pratu043
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pls help.
lgbasallote
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welcome to logarithms!
pratu043
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Yeah thanks!
waterineyes
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I guess you can solve this without using Logs..
pratu043
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But I want to know how to do it with log.
waterineyes
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Okay then all here will explain you how to do this...
moha_10
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15^(x-3)+(2x-8)=225
now we take ln
(3x-11)15= ln225
pratu043
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What is In?
And how did you get 15^(x-3)+(2x-8)=225?
moha_10
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15*ln(3x-11)=ln 225
moha_10
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by multi 3 into 5 to get 15
and the exponent it's added
pratu043
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But how can you just multiply the base and add the exponents?
waterineyes
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Use the following formulas of Logs:
\[\large Log(a \times b) = Log(a) + Log(b)\]
\[\large Log(a)^b = b.Log(a)\]
moha_10
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before we take the logarithem
moha_10
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it's from exponent experepties
pratu043
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Why did you delete that @dpaInc?
dpaInc
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because i don't wanna take away from these guys awesome explanations....:)
waterineyes
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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..
waterineyes
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Use the first formula only...
pratu043
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\[\log(5^{x-3}) . \log(3^{2x-8}) = \log 225 \]
waterineyes
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Now Check my first formula carefully...
moha_10
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@pratu043 trust take my answer
moha_10
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like what waterineyes did now
pratu043
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That's what I've done here right @waterineyes?
waterineyes
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I give you an example:
\[\large Log(2 \times 3) = Log(2) + Log(3)\]
Go by this procedure...
pratu043
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I've already done that. So now we should use the second one?
waterineyes
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You have done it wrongly my friend...
waterineyes
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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???
pratu043
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Oops.
\[\log(5^{x-3}) + \log(3^{2x-8}) = \log225\]
waterineyes
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Yes now you are right...
Now 225 is square of what number???
pratu043
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15
waterineyes
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So can you write 225 as \(15^2\) on right hand side.??
If yes then do it..
pratu043
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\[\log(5^{x-3}) + \log(3^{2x-8}) = \log15^{2}\]
waterineyes
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Now use the power rule of Logarithms that is the Second Formula that I have given above..
Can you use that??
waterineyes
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I give you an example of that too:
\[\large Log(x)^4 = 4.Log(x)\]
pratu043
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\[(x-3)\log5 + (2x-8)\log3 = 2\log15\]
waterineyes
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Can you write 15 as 3*5??
pratu043
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Yes.
\[(x-3)\log5 + (2x-8)\log3 = 2\log(5 * 3)\]
waterineyes
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Then use it and use the first formula again on right hand side...
pratu043
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\[(x-3)\log5 + (2x-8)\log3 = 2\log5 + 2\log3\]
waterineyes
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Now what you have to do is to compare the coefficients of \(Log3\) and \(Log5\) both the sides can you do that??
waterineyes
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Tell me the coefficient of Log3 on Left hand side and right hand side...
pratu043
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On LHS its 2x - 8 and on RHS its 2.
waterineyes
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Just equate them and find x from it..
waterineyes
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Similarly equate the coefficients of log5 and then also find x you will get the same x in both the case...
waterineyes
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No no...
pratu043
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\[(2x-8)\log3 = 2\log3\]
waterineyes
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I said to equate the coefficients buddy..
Coefficients are 2x - 8 and 2..
you should equate this only...
pratu043
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So you should just do 2x-8 = 2?
pratu043
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x = 5
waterineyes
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See, if :
\[\large xlog3 = ylog3\]
implies x = y.
Getting??
pratu043
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Yes!!
waterineyes
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Yes you have found x = 5 that is absolutely right...
pratu043
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Thanks a lot!!
waterineyes
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Now equate the coefficients of log5..
pratu043
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x - 3 = 2
x = 2 + 3
= 5
waterineyes
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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??