# Logarithm question

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Bigflakes

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#1

mqb2766

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Bigflakes

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#3

(Original post by

Try taking logs.

**mqb2766**)Try taking logs.

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mqb2766

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#4

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#4

(Original post by

So log1+x/100(1.25) = 12

**Bigflakes**)So log1+x/100(1.25) = 12

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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Bigflakes

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#5

(Original post by

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.

**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.

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mqb2766

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#6

(Original post by

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 ?

**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 ?

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

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.

**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.

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#8

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#8

(Original post by

So log(1+(x/100)) = log(1.25)/(1/12) then 1+(x/100) = 1.162920156 then solve for x so x = 16.29201561 ?

**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 ?

Why do you divide by 1/12 at the start? Again, its close but wrong.

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#9

(Original post by

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.

**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.

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#10

(Original post by

Do I divide by 12 ?

**Bigflakes**)Do I divide by 12 ?

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

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.

**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.

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#12

(Original post by

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

**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

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

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.

**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.

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#14

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#14

(Original post by

Okay so I got 1.8769, does this seem reasonable? I did 10^x after I did log(1.25)/12

**Bigflakes**)Okay so I got 1.8769, does this seem reasonable? I did 10^x after I did log(1.25)/12

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

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.

**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.

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#16

(Original post by

Okay thank you, could you also help me solve ln(x) = 3-ln(2x)

**Bigflakes**)Okay thank you, could you also help me solve ln(x) = 3-ln(2x)

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#17

(Original post by

Can you post what youve tried. There are a few similar first steps.

**mqb2766**)Can you post what youve tried. There are a few similar first steps.

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#18

(Original post by

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?

**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?

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

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?

**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?

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#20

(Original post by

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

**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

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