# C2 Log question

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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)

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)

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#6

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 :/

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(Original post by

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 :/

**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 :/

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#8

**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 :/

0

reply

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)

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)

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#10

(Original post by

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)

**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)

alog(b)=log(b^a)

log(a)+log(b)=log(ab)

log(a)-log(b)=log(a/b)

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(Original post by

Use your laws of logs:

alog(b)=log(b^a)

log(a)+log(b)=log(ab)

log(a)-log(b)=log(a/b)

**CJG21**)Use your laws of logs:

alog(b)=log(b^a)

log(a)+log(b)=log(ab)

log(a)-log(b)=log(a/b)

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#12

(Original post by

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

**Sam160**)I have an answer for the first question which i know is wrong from the 2nd which is the problem

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(Original post by

What is your answer to the first?

**CJG21**)What is your answer to the first?

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#14

(Original post by

I got to log10[(x+4)/x^2] + log10(x+16) but going wrong after

**Sam160**)I got to log10[(x+4)/x^2] + log10(x+16) but going wrong after

Spoiler:

Show

log[(x+4)(x+16)/x^2]

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