DaQuestioner
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5^x=6^(x-1)

Solve the equation using logarithms.
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mqb2766
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(Original post by DaQuestioner)
5^x=6^(x-1)

Solve the equation using logarithms.
So what are you stuck with. It's just the normal log rule
log(a^b) = b*log(a)
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DaQuestioner
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(Original post by mqb2766)
So what are you stuck with. It's just the normal log rule
log(a^b) = b*log(a)
Yes but how do you use that to solve this; I am confused about there being an x on both sides :/
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ThiagoBrigido
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Have you tried to look as 6^(x-1), which is equal to 6^x/6^-1? Thereafter isolate the terms ^x to one side and try again.
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mqb2766
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(Original post by DaQuestioner)
Yes but how do you use that to solve this; I am confused about there being an x on both sides :/
As above or just take logs and pop the xs to one side as usual ? Show your attempt?
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ghostwalker
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(Original post by ThiagoBrigido)
Have you tried to look as 6^(x-1), which is equal to 6^x/6^-1?
Close, but no cigar.


6^{x-1}=6^x6^{-1}=6^x/6
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DaQuestioner
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So far, I have done:

log 5 5^x = log 5 6^(x-1)
xlog 5 5 = (x-1)log 5 6

But I don't know what to do next
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mqb2766
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(Original post by DaQuestioner)
So far, I have done:

log 5 5^x = log 5 6^(x-1)
xlog 5 5 = (x-1)log 5 6

But I don't know what to do next
Expand the right
Take the terms involving x to the left and factorise
Then solve

Note, you don't have to use base 5?
Last edited by mqb2766; 1 month ago
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hhc2
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(Original post by DaQuestioner)
So far, I have done:

log 5 5^x = log 5 6^(x-1)
xlog 5 5 = (x-1)log 5 6

But I don't know what to do next
(Original post by DaQuestioner)
5^x=6^(x-1)

Solve the equation using logarithms.
xlog(10)5=x-1 log(10) 6
(x-1)/x=(log10 5)/(log10 6)
1-(x^-1) = (log10 5)/(log10 6)
1- (log10 5)/(log10 6) = x^-1
then just
1/ (1- (log10 5)/(log10 6)) = x


right?
Last edited by hhc2; 1 month ago
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hhc2
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(Original post by hhc2)
xlog(10)5=x-1 log(10) 6
(x-1)/x=(log10 5)/(log10 6)
1-(x^-1) = (log10 5)/(log10 6)
1- (log10 5)/(log10 6) = x^-1
then just
1/ (1- (log10 5)/(log10 6)) = x


right?
that seems wrong acc
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ThiagoBrigido
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(Original post by ghostwalker)
Close, but no cigar.


6^{x-1}=6^x6^{-1}=6^x/6
HAHHAHA. What a silly mistake, that would just confuse him even more. Thanks for pointing that out.
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mqb2766
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(Original post by hhc2)
xlog(10)5=x-1 log(10) 6
(x-1)/x=(log10 5)/(log10 6)
1-(x^-1) = (log10 5)/(log10 6)
1- (log10 5)/(log10 6) = x^-1
then just
1/ (1- (log10 5)/(log10 6)) = x


right?
Eh?
x*log(5) = x*log(6) - log(6)
Do the steps in the previous post. That's one done.
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mqb2766
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(Original post by ashpritgcse2020)
hello hopefully i can help , im a gcse student but flip.

5^x = 6^x-1

powers are c and d and number/ba
pls delete. Hints not solutions
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DaQuestioner
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(Original post by mqb2766)
Expand the right
Take the terms involving x to the left and factorise
Then solve

Note, you don't have to use base 5?
I have been using base 5 so far, and this is where I have got to since after he last few steps:
xlog 5 5 = xlog 5 6 - log 5 6 (although have I made a mistake here; not sure )
log 5 6 = xlog 5 6 - xlog 5 5
log 5 6 = log 5 6^x - log 5 5^x
log 5 6 = log 5 (6^x/5^x)
6^x/5^x = 6
and then I'm lost again
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ashpritgcse2020
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(Original post by DaQuestioner)
I have been using base 5 so far, and this is where I have got to since after he last few steps:
xlog 5 5 = xlog 5 6 - log 5 6 (although have I made a mistake here; not sure )
log 5 6 = xlog 5 6 - xlog 5 5
log 5 6 = log 5 6^x - log 5 5^x
log 5 6 = log 5 (6^x/5^x)
6^x/5^x = 6
and then I'm lost again
you dont need to do that -> look at mine its easier> you dont need to use the full logarithm
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ashpritgcse2020
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(Original post by mqb2766)
pls delete. Hints not solutions
solutions are hints because he can understand and do a similar question. dont you dare use imperatives
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DaQuestioner
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(Original post by ashpritgcse2020)
hello hopefully i can help , im a gcse student but flip.

5^x = 6^x-1

powers are c and d and number/base is and b in the formula: ie ; c is for the power of a and d is the power for b

a^c = b^d

c(log a) = d(log b)

x(log5) = (x-1)(log6)
open the brackets on the right side
x(log5) = x(log6) - (log6)
rearrange the equation to get all x values on 1 side
log6 = x(log6) - x(log5)
then take x out
log6 = x(log6-log5)
rearrange to make x the subject
(log6) /(log6 - log5) = x
x should be equal to 9.827 to 3 s.f but idk cus im a gcse student.
Thank you! I understand now Just didn't see think of that 'taking out the x' thing.
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DaQuestioner
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(Original post by ashpritgcse2020)
solutions are hints because he can understand and do a similar question. dont you dare use imperatives
(Original post by ThiagoBrigido)
HAHHAHA. What a silly mistake, that would just confuse him even more. Thanks for pointing that out.
Yo guys it's 2020; girls ask for maths help too yk :bhangra:
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mqb2766
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(Original post by ashpritgcse2020)
solutions are hints because he can understand and do a similar question. dont you dare use imperatives
It's the forum rules. Pls read them.
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ashpritgcse2020
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(Original post by DaQuestioner)
Yo guys it's 2020; girls ask for maths help too yk :bhangra:
whoops, sorry my bad
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