Easy logarithm question but im stupid Watch

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geoffhig
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#1
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#1
2^x+1=5^x

solve for x
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Spenceman_
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#2
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Did you end up with having to solve this as part of a larger question? Or otherwise where it is from?
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yusufu
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is that 2^(x+1) or (2^x) + 1
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Jonny W
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#4
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I interpret your equation as 2^(x + 1) = 5^x (not (2^x) + 1 = 5^x).

Taking logs of both sides and using the rule ln(a^b) = b ln(a) gives

(x + 1)ln(2) = x ln(5)
x(ln(5) - ln(2)) = ln(2)

x
= ln(2)/(ln(5) - ln(2))
= 0.756
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amo1
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(Original post by geoffhig)
2^x+1=5^x

solve for x
xln2 + ln1 = xln5
ln 1 =0

so x=0

hmm that dont seem right
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amo1
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unless its 2x+1 ?
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m:)ckel
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I'm assuming it is 2x+1, but in any case, you can't do what you did in your previous post, i.e. split the terms in the 'ln' up. It should be:
ln(2x+1) = ln5x, if you were to interpret it as that...
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Mr Laminator
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(Original post by Jonny W)
I interpret your equation as 2^(x + 1) = 5^x (not (2^x) + 1 = 5^x).

Taking logs of both sides and using the rule ln(a^b) = b ln(a) gives

(x + 1)ln(2) = x ln(5)
x(ln(5) - ln(2)) = ln(2)

x
= ln(2)/(ln(5) - ln(2))
= 0.756
I like this answer.
And I wouldn't like the question if it was indeed (2^x) + 1 = (5^x).
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#9
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(Original post by amo1)
xln2 + ln1 = xln5
ln 1 =0

so x=0

hmm that dont seem right
cant do that cos they aint products
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#10
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#10
just to make it interesting - how would you solve this question if it was:

2^x -5^x =-1 ???

been thinking about it and cant see a log way of doing it, nor any other way? could you do some sort of approximation from a series or something :confused:

pk
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Aitch
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(Original post by Mr Laminator)
I like this answer.
And I wouldn't like the question if it was indeed (2^x) + 1 = (5^x).
It looks as though it should be soluble, doesn't it!

Especially expressed as

2x +1x =5x

Aitch
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geoffhig
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#12
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#12
yes it is 2^(x+1)
its from the C1/C2 book chapter 18D question 2.
The answer is 0.756 thanks for the method. When you take logs from both sides can you use either log base ten or the ln thing?
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m:)ckel
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(Original post by geoffhig)
When you take logs from both sides can you use either log base ten or the ln thing?
Yeah, any base will do. Just make sure you use the same base throughout (obviously).
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#14
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(Original post by Aitch)
It looks as though it should be soluble, doesn't it!

Especially expressed as

2x +1x =5x

Aitch
looks a bit like fermats last theorum :eek: :rolleyes:

and from the idea of the proof i thought Wiles proved that it had no solutions...hmmmm..oh yes - that was no integer solutions - right?
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Spenceman_
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#15
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yeh if it was

2^x + 1 = 5^x

there is a solution for 0 < x < 1 .... you can see there would be from the graphs. I went and found it numerically because I thought thats what the question was but it was nothing special.
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geoffhig
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#16
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#16
1-2log(base2)x=log(base2)(5x-12)

i keep ending up with 5x³-12x²=2 which is wrong because the answers say 4,6
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Vijay1
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#17
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(Original post by Spenceman_)
yeh if it was

2^x + 1 = 5^x

there is a solution for 0 < x < 1 .... you can see there would be from the graphs. I went and found it numerically because I thought thats what the question was but it was nothing special.
the solution for that would be round about 0.555 :rolleyes:
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Jonny W
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#18
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#18
(Original post by geoffhig)
1-2log(base2)x=log(base2)(5x-12)

i keep ending up with 5x3-12x2=2 which is wrong because the answers say 4,6
I think the equation should be

-1 + 2log2(x) = log2(5x - 12)

Then

(1/2)x^2 = 5x - 12
x^2 - 10x + 24
(x - 4)(x - 6) = 0
x = 4 or 6
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Mr Laminator
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#19
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(Original post by Spenceman_)
yeh if it was

2^x + 1 = 5^x

there is a solution for 0 < x < 1 .... you can see there would be from the graphs. I went and found it numerically because I thought thats what the question was but it was nothing special.
Please would u be able to post your numerical solution?
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Spenceman_
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#20
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#20
2x + 1 = 5x

ln(2x + 1) = ln5x

ln(2x + 1) = xln5


x = [ln(2x + 1)]/ln5

Setting up an iterative process:-

xn+1 = [ln(2xn + 1)]/ln5

Start with x1 = 1

Iteration converges to x = 0.563895524
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