# Easy logarithm question but im stupidWatch

This discussion is closed.
#1
2^x+1=5^x

solve for x
0
13 years ago
#2
Did you end up with having to solve this as part of a larger question? Or otherwise where it is from?
0
13 years ago
#3
is that 2^(x+1) or (2^x) + 1
0
13 years ago
#4
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
0
13 years ago
#5
(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
0
13 years ago
#6
unless its 2x+1 ?
0
13 years ago
#7
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...
0
13 years ago
#8
(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
And I wouldn't like the question if it was indeed (2^x) + 1 = (5^x).
0
13 years ago
#9
(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
0
13 years ago
#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

pk
0
13 years ago
#11
(Original post by Mr Laminator)
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
0
#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?
0
13 years ago
#13
(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).
0
13 years ago
#14
(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

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?
0
13 years ago
#15
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.
0
#16
1-2log(base2)x=log(base2)(5x-12)

i keep ending up with 5xÂ³-12xÂ²=2 which is wrong because the answers say 4,6
0
13 years ago
#17
(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
0
13 years ago
#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
0
13 years ago
#19
(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.
0
13 years ago
#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

Iteration converges to x = 0.563895524
0
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