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# logarithms watch

1. How would I go about solving this problem:
5000x = 2^(x-1)

Thanks
2. 5000x = 2^(x-1)
5000x = (x-1)(x-1)
5000x = x^2 - x - x + 1 = 0
x^2 -5002x + 1 = 0
(x - 2501)^2 - 2501^2 + 1 = 0 {complete the square}
(x - 2501)^2 = 6255000
x - 2501 = +/-sqrt(6255000)
x = 2501 +/- 2500.9998
x= 5001.9998 OR x = 1/5000
i think.
3. (Original post by Revelation)
5000x = 2^(x-1)
5000x = (x-1)(x-1)
I don't see how you've gone from the first line to the second :/.
4. (Original post by Nylex)
I don't see how you've gone from the first line to the second :/.
quite!
You could say that 2^(x-1) = 2^x / 2 thats about it!
How would I go about solving this problem:
5000x = 2^(x-1)

Thanks
Have you tried doing it by trial and error?
I get x=17.4 (approx) - how accurate does it need to be?
6. On a similar vein, see this topic:

I think that this equation is probably trancendental like ln(x)/x=c, so it will not be possible to solve algebraicly. Try a numerical approach.
How would I go about solving this problem:
5000x = 2^(x-1)

Thanks
log5000+logx=(x-1)log2
log5000+logx=xlog2-log1
log5001=xlog2-logx

You can do the rest.
8. (Original post by Bhaal85)
log5000+logx=(x-1)log2
log5000+logx=xlog2-log1
log5001=xlog2-logx

You can do the rest.
J/k.

This is actually quite a *****.

5000X=2^(X-1)
log5000+logX=(X-1)log2
log5000+logX=Xlog2-log2
log5000+log2=Xlog2-logX
log(5000*2)=Xlog2-log2
log(10000)=Xlog2-log2

Now I'm stuck.
9. (Original post by Bhaal85)
J/k.

This is actually quite a *****.

5000X=2^(X-1)
log5000+logX=(X-1)log2
log5000+logX=Xlog2-log2
log5000+log2=Xlog2-logX
log(5000*2)=Xlog2-log2
log(10000)=Xlog2-log2

Now I'm stuck.
log10000=Xlog2/log2 therefire X = log10000????
10. Numerical approach solutuion.
Attached Images

11. thnx foir replies. i tried doin it algebraically but kept hitting dead ends. thanks for explaining y it can't be solved algebraically too.

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Updated: March 16, 2004
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