Logarithmic Problem

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dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
If, M^(x-1)=N^x, then x equals:

I didn't know how to start this so I looked at the explaination and it said the first step was:

M^(x-1)=N^x (Take the logarithm of both sides)
(x-1)log(M)=xlog(N)

The problem I'm having is that I don't understand how M^(x-1)=(x-1)log(M) and N^x=xlog(N).

Contradiction
Let's just use N^x=xlog(N) for example. Say N=1 and x=2:

N^x=xlog(N)
1^2=2log(1)
1=2(0)
1=0

So N^x doesn't = xlog(N).

If someone could explain where my contridiction is wrong and why this first step works, that would be great! :D
 
They aren't equal. If you saw
5x = 10 and
x = 2 would you say
2 doesn't = 10 so there is a contradiction???
Same thing here. You are replacing both sides by an equivalent expression.
3log(100)=6
log(100^3)=6
They are equivalent expressions.
As long as you do the same thing to both sides (divide by 5 or take the logs) it works.
 
Hello, dagr8est!

If you're familiar with logarithms at all, you should understand . . .

]If, M^(x-1)=N^x, then x equals:

I didn't know how to start this so I looked at the explaination and it said the first step was:

M^(x-1)=N^x (Take the logarithm of both sides)
. . . . . . . . . . . . . . . . . . <--- Did you omit a step?
(x-1)log(M)=xlog(N)

The problem I'm having is that I don't understand how M^(x-1)=(x-1)log(M) and N^x=xlog(N).
Click to expand...
As Gene pointed out, those are <u>not</u> equal . . . there is a step missing.

We have: . M<sup>x-1</sup> . = . N<sup>x</sup>

Take logs: . log(M<sup>x-1</sup>) . = . log(N<sup>x</sup>)

There is a property of logs that allows us to "bring down" the exponent . . . remember?

And that's why we have: . (x-1) log M . = . x log N

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

What you're doing goes something like this . . .

We have: . 2x .= .6
. . . . . . . . . . ↓ . . . ↓ . . . (We invisibly and <u>secretly</u> divide both sides by 2)
And then: . .x . = .3

And you're thinking: "I don't understand how 2x = x and how 6 = 3."
 
Okay I understand it now. I just got confused because they didn't show log(M^(x-1))=log(N^x) which caused me to misinterpret what they were doing. Thanks for the help.
 
hello dagr8est

M^x-1=n^x

logm^x-1=logn^x

(x-1)logm=x(logn)

xlogm-x=logn-logm

x(logm-1)=logn-logm

x=logn-logm/logm-1
 
Hate to disagree with the new kid (dispite his careless ()s), but what I get is:
(x-1)logm=x(logn)
x*log(m)-log(m)=x*log(n)
x(log(m)-log(n)) = log(m)
x=log(m)/(log(m)-log(n)) =

log(m)/log(m/n)
 
Gene is correct. The answer is:

M^(x-1)=N^x
log[M^(x-1)]=logN^x
(x-1)logM=xlogN
xlogM-logM=xlogN
xlogM-xlogN=logM
x(logM-logN)=logM
x=logM/(logM-logN)
 
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