Logarithm question

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#1
Can anyone point me in the right direction to solve x:

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1 month ago
#2
Try taking logs.
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#3
(Original post by mqb2766)
Try taking logs.
So log1+x/100(1.25) = 12
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1 month ago
#4
(Original post by Bigflakes)
So log1+x/100(1.25) = 12
If you have
log(a^b) = log(c)
you should get
log(a) = log(c)/b
then inverse logs and solve for x.

You could have taken the 1/12 root on each side, then taken logs. Then ... it'll give the same result.

Note its possible to estimate x directly as for small values compound interest is basically simple interest.
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#5
(Original post by mqb2766)
If you have
log(a^b) = log(c)
you should get
log(a) = log(c)/b
then inverse logs and solve for x.

You could have taken the 1/12 root on each side, then taken logs. Then ... it'll give the same result.

Note its possible to estimate x directly as for small values compound interest is basically simple interest.
Sorry, I’m really confused. So log(1+(x/100)) = (log(1.25))/(log(12)) , I don’t see how this will help unless I’m suppose to work out the right hand side that subtract 1 and time by 100 to get x ?
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1 month ago
#6
(Original post by Bigflakes)
Sorry, I’m really confused. So log(1+(x/100)) = (log(1.25))/(log(12)) , I don’t see how this will help unless I’m suppose to work out the right hand side that subtract 1 and time by 100 to get x ?
Its not log(12) on the right hand side, see previous post or the index rules.
But the right hand side is just a number, so evaulate it then take inverse logs of both sides and solve for x.
Last edited by mqb2766; 1 month ago
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#7
(Original post by mqb2766)
Its not log(12) on the right hand side, see previous post or the index rules.
But the right hand side is just a number, so evaulate it then take inverse logs of both sides and solve for x.
So log(1+(x/100)) = log(1.25)/(1/12) then 1+(x/100) = 1.162920156 then solve for x so x = 16.29201561 ?
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1 month ago
#8
(Original post by Bigflakes)
So log(1+(x/100)) = log(1.25)/(1/12) then 1+(x/100) = 1.162920156 then solve for x so x = 16.29201561 ?
First that cant be the answer. Check by subbing in your calculator or note that to get 25% using simple interest over 12 years youd have to have 2% a year, so x~2.
Why do you divide by 1/12 at the start? Again, its close but wrong.
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#9
(Original post by mqb2766)
First that cant be the answer. Check by subbing in your calculator or note that to get 25% using simple interest over 12 years youd have to have 2% a year, so x~2.
Why do you divide by 1/12 at the start? Again, its close but wrong.
Do I divide by 12 ?
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1 month ago
#10
(Original post by Bigflakes)
Do I divide by 12 ?
It shouldn't even be a question if you're applying the index laws properly, but yes.
When you get your answer, try validating by subbing the answer into the original expression to check.
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#11
(Original post by mqb2766)
It shouldn't even be a question if you're applying the index laws properly, but yes.
When you get your answer, try validating by subbing the answer into the original expression to check.
Okay, but when I did log(1.25)/12 , I subtracted 1 and then multiplied by 100, I got -99.2 could you explain what I did wrong
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1 month ago
#12
(Original post by Bigflakes)
Okay, but when I did log(1.25)/12 , I subtracted 1 and then multiplied by 100, I got -99.2 could you explain what I did wrong
Youre only writing down part of the working. The equation is
log(1+(x/100)) = log(1.25)/12
So evaluate the right hand side and take inverse logs, then solve for x.
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#13
(Original post by mqb2766)
Youre only writing down part of the working. The equation is
log(1+(x/100)) = log(1.25)/12
So evaluate the right hand side and take inverse logs, then solve for x.
Okay so I got 1.8769, does this seem reasonable? I did 10^x after I did log(1.25)/12
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1 month ago
#14
(Original post by Bigflakes)
Okay so I got 1.8769, does this seem reasonable? I did 10^x after I did log(1.25)/12
If your log is base 10, then that should be good. But sub into your original expression/calculator if you want to verify.

Note the previous simple estimate for x in the previous post x~2. Also, a simpler (non log) method would have been to root 1.25 twice and then cube root it again and then solve for x. As this would be power 1/12.
Last edited by mqb2766; 1 month ago
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#15
(Original post by mqb2766)
If your log is base 10, then that should be good. But sub into your original expression/calculator if you want to verify.
Note the previous simple estimate for x in the previous post x~2.
Okay thank you, could you also help me solve ln(x) = 3-ln(2x)
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1 month ago
#16
(Original post by Bigflakes)
Okay thank you, could you also help me solve ln(x) = 3-ln(2x)
Can you post what youve tried. There are a few similar first steps.
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#17
(Original post by mqb2766)
Can you post what youve tried. There are a few similar first steps.
Well I moved the negative to the left , then combined them to get 2log(2x)=3 so then I did 3/2 and put it into 10^x, the finally divided by 2 to get x alone?
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1 month ago
#18
(Original post by Bigflakes)
Well I moved the negative to the left , then combined them to get 2log(2x)=3 so then I did 3/2 and put it into 10^x, the finally divided by 2 to get x alone?
How did you combine them and get the 2 multiplier in front of the log?
Im guessing youre making simple mistakes as you're not being methodical enough.
Also is ln() in the question the natural log (base e) or base 10?
Last edited by mqb2766; 1 month ago
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#19
(Original post by mqb2766)
How did you combine them and get the 2 multiplier in front of the log?
Im guessing youre making simple mistakes as you're not being methodical enough.
Also is ln() in the question the natural log (base e) or base 10?
So I added the ln(2x) to both sides and then I got ln(x) + ln(2x) = 3 so ln(2x^2)=3 and then put the 2 infront
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1 month ago
#20
(Original post by Bigflakes)
So I added the ln(2x) to both sides and then I got ln(x) + ln(2x) = 3 so ln(2x^2)=3 and then put the 2 infront
2ln(2x) = ln(4x^2)
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