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C2 log question

in question 6c. my steps are fine until the last step:
N log_7/8(7/8)< log_7/81/320
N < 43.2 (3sf)

However the correct answer is N > 43.2.
Can someone explain why?

Reply 1

Zacken

Original post by alesha98

in question 6c. my steps are fine until the last step:
N log_7/8(7/8)< log_7/81/320
N < 43.2 (3sf)

However the correct answer is N > 43.2.
Can someone explain why?

You're dividing by a negatie number.

Reply 2

Notnek

Original post by alesha98

in question 6c. my steps are fine until the last step:
N log_7/8(7/8)< log_7/81/320
N < 43.2 (3sf)

However the correct answer is N > 43.2.
Can someone explain why?

Your penultimate line is also incorrect.

Can you post your working in full?

Reply 3

alesha98

Original post by Zacken

You're dividing by a negatie number.

my full working for part c is:
160 - (20(1-(7/8)^N)) / 1-7/8 < 0.5
20 - 20(1-(7/8)^N) < 1/16
320 - 320(1-(7/8)^N) < 1
320(1-(7/8)^N) > 319
1-(7/8)^N > 319/320
-(7/8)^N > -1/320
(7/8)^N < 1/320
N log_7/8(7/8) < log_7/81/320
N < 43.2 (3sf)

May you tell me where exactly I got wrong please?

Reply 4

alesha98

Original post by notnek

Your penultimate line is also incorrect.

Can you post your working in full?

yes i just did

Reply 5

Zacken

Original post by alesha98

yes i just did

Do you understand what it means to multiply and divide an inequality by a negative number?

I'll let Notnek take over, I'm not home right now and can't check your working thoroughly.

(edited 8 years ago)

Reply 6

the bear

remember that some logarithms are negative ?

so if you divide by a logarithm you may need to flip the arrow....

Reply 7

alesha98

Original post by the bear

remember that some logarithms are negative ?

so if you divide by a logarithm you may need to flip the arrow....

I dont know anything about that, can you explain it please?

Reply 8

the bear

Original post by alesha98

I dont know anything about that, can you explain it please?

for instance log0.8 is a negative number ( using base 10 )

so if you have

X(log0.8) > 15

X < 15/log0.8

if instead you have

X(log8) > 15

X > 15/log8

Reply 9

alesha98

Original post by the bear

for instance log0.8 is a negative number ( using base 10 )

so if you have

X(log0.8) > 15

X < 15/log0.8

if instead you have

X(log8) > 15

X > 15/log8

thanks. So if i log a number that is smaller than 1, then it is a negative number ?

Reply 10

the bear

Original post by alesha98

thanks. So if i log a number that is smaller than 1, then it is a negative number ?

yes that is true for any base.

Reply 11

alesha98

Original post by the bear

for instance log0.8 is a negative number ( using base 10 )

so if you have

X(log0.8) > 15

X < 15/log0.8

if instead you have

X(log8) > 15

X > 15/log8

but from what i did, N log_7/8(7/8) < log_7/81/320, log_7/8(7/8) is 1, so why do i need to flip the inequality?

Reply 12

alesha98

Original post by notnek

Your penultimate line is also incorrect.

Can you post your working in full?

what happen if I log with a base that is smaller than 1?

Reply 13

Notnek

Original post by alesha98

but from what i did, N log_7/8(7/8) < log_7/81/320, log_7/8(7/8) is 1, so why do i need to flip the inequality?

Taking logs of base

a

where

0<a<1

reverses the inequality.

If you take logs base 10 instead then things are clearer:

\displaystyle \left(\frac{7}{8}\right)^N < \frac{1}{320}

\displaystyle N \log_{10} \left(\frac{7}{8}\right)< \log_{10} \left(\frac{1}{320}\right)

Now

\log_{10} \left(\frac{7}{8}\right) < 0

so we need to reverse the inequality when dividing:

\displaystyle N > \frac{ \log_{10} \left(\frac{1}{320}\right)}{\log_{10} \left(\frac{7}{8}\right)}

Reply 14

alesha98

Original post by notnek

Taking logs of base

a

where

0<a<1

reverses the inequality.

thankyou so much

Reply 15

Notnek

Original post by alesha98

Taking logs of base

a

where

0<a<1

reverses the inequality.
To explain this a bit further :

\displaystyle \log_{\frac{7}{8}} x

is a strictly decreasing function i.e. it looks like this :

So e.g.

2 < 3

but

\displaystyle \log_{\frac{7}{8}} 2 > \log_{\frac{7}{8}} 3

as you can see in the graph.

\log_{10} x

is strictly increasing so you don't need to reverse the inequality sign when taking logs base 10.

(edited 8 years ago)

Reply 16

Zacken

Original post by alesha98

thankyou so much

I'd suggest always using log base 10 or the natural log for these questions. Don't do things like logs with fractional bases or whatever, it's just complicating your life.

Reply 17

Notnek

Original post by Zacken

I'd suggest always using log base 10 or the natural log for these questions. Don't do things like logs with fractional bases or whatever, it's just complicating your life.

I agree with this.

Or you could treat a log inequality as an equation and only deal with the inequality sign at the end, similar to how one may tackle a quadratic inequality.