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C2 Log question

Stuck with this:

Find the solution

(2^x)(2^x+1) = 10

Thanks

Slightly stuck with another question too:

Write log10(x+4) - 2log10(x)+log10(x+16) as a single log

Without a calculator show x=4 is a root log10(x+4) - 2log10(x)+log10(x+16)
(edited 10 years ago)
Reply 1
Original post by Sam160
Stuck with this:

Find the solution

(2^x)(2^x+1) = 10

Thanks


Can you clarify which of these is the problem

(2x)(2x+1)=10(2^x)(2^x+1) = 10

or

(2x)(2x+1)=10(2^x)(2^{x+1}) = 10
(x^a)(x^b) = x^(a+b)
Original post by Sam160
Stuck with this:

Find the solution

(2^x)(2^x+1) = 10

Thanks


Original post by StrangeBanana
(x^a)(x^b) = x^(a+b)


This :smile:
Reply 4
Original post by TenOfThem
Can you clarify which of these is the problem

(2x)(2x+1)=10(2^x)(2^x+1) = 10

or

(2x)(2x+1)=10(2^x)(2^{x+1}) = 10


The 2nd one, thanks for the help guys
Reply 5
Original post by Sam160
Stuck with this:

Find the solution

(2^x)(2^x+1) = 10

Thanks


If it is

I got x = 1.16 (3sf)

Log all sides so

Log (2^x) (Log 2^x+1) = Log 10

Log (2^x * 2^x+1 ) = Log 10

Log (2^2x +1) = 1 (Log 10 = 1)

(2x+1)Log 2 = 1
(2x + 1) = 1 / Log 2
rearranging gives x = 1.160964047 which is 1.16(3sf)


NOTE: I have just recently learnt this topic so please confirm to me whether my answer is right or if you dont have the mark scheme. Other people should attempt the question to confirm my answer or disapprove :/
Reply 6
Original post by AhmedDavid
If it is

I got x = 1.16 (3sf)

Log all sides so

Log (2^x) (Log 2^x+1) = Log 10

Log (2^x * 2^x+1 ) = Log 10

Log (2^2x +1) = 1 (Log 10 = 1)

(2x+1)Log 2 = 1
(2x + 1) = 1 / Log 2
rearranging gives x = 1.160964047 which is 1.16(3sf)


NOTE: I have just recently learnt this topic so please confirm to me whether my answer is right or if you dont have the mark scheme. Other people should attempt the question to confirm my answer or disapprove :/


That is what i have got now too, thanks for your help!
Reply 7
Original post by AhmedDavid
If it is

I got x = 1.16 (3sf)

Log all sides so

Log (2^x) (Log 2^x+1) = Log 10

Log (2^x * 2^x+1 ) = Log 10

Log (2^2x +1) = 1 (Log 10 = 1)

(2x+1)Log 2 = 1
(2x + 1) = 1 / Log 2
rearranging gives x = 1.160964047 which is 1.16(3sf)


NOTE: I have just recently learnt this topic so please confirm to me whether my answer is right or if you dont have the mark scheme. Other people should attempt the question to confirm my answer or disapprove :/


Your method is correct and just remember to ALWAYS work in a logical and methodical manner. Many people make mistakes when they try to do too much in their heads!
Reply 8
Slightly stuck with another question too:

Write log10(x+4) - 2log10(x)+log10(x+16) as a single log

Without a calculator show x=4 is a root log10(x+4) - 2log10(x)+log10(x+16)
Original post by Sam160
Slightly stuck with another question too:

Write log10(x+4) - 2log10(x)+log10(x+16) as a single log

Without a calculator show x=4 is a root log10(x+4) - 2log10(x)+log10(x+16)


Use your laws of logs:

alog(b)=log(b^a)
log(a)+log(b)=log(ab)
log(a)-log(b)=log(a/b)
Reply 10
Original post by CJG21
Use your laws of logs:

alog(b)=log(b^a)
log(a)+log(b)=log(ab)
log(a)-log(b)=log(a/b)


I have an answer for the first question which i know is wrong from the 2nd which is the problem
Original post by Sam160
I have an answer for the first question which i know is wrong from the 2nd which is the problem


What is your answer to the first?
Reply 12
Original post by CJG21
What is your answer to the first?


Nevermind i've worked out both
(edited 10 years ago)
Original post by Sam160
I got to log10[(x+4)/x^2] + log10(x+16) but going wrong after


Well you can use your log laws to simplify that to one log.

Spoiler

Original post by Sam160
Nevermind i've worked out both


Excellent. :grin:

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