Logarithms and exponential functions

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Shas72
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#1
Report Thread starter 8 months ago
#1
4^x=7(3^x) solve giving your answer to 3sf.
I dont understand
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Shas72
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#2
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(Original post by Shas72)
4^x=7(3^x) solve giving your answer to 3sf.
I dont understand
So for 4^x I write as 2^x+1
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David Getling
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(Original post by Shas72)
So for 4^x I write as 2^x+1
No. Try taking the logarithm of each side.
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Shas72
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#4
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I did but then x gets cancelled
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mqb2766
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#5
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(Original post by Shas72)
I did but then x gets cancelled
Can you post your working? Taking logs (any base) should get the answer.
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Shas72
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#6
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#6
so if you do xlog4= 7× x log 3
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mqb2766
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#7
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(Original post by Shas72)
so if you do xlog4= 7× x log 3
Nearly. log(A*B) = ?
Its the main property for logs. You've done the exponent bit correctly. So
log(7*3^x) = ?
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Shas72
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#8
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so x gets cancelled
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Shas72
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#9
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#9
oh ok so it will be multiplication law
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mqb2766
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#10
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(Original post by Shas72)
oh ok so it will be multiplication law
:-)
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Shas72
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#11
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#11
log 7+xlog3
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Shas72
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#12
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#12
thank you so so much. I got it
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RDKGames
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#13
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#13
You could also divide both sides by 3^x to get

4^x / 3^x = 7

The LHS is simply (4/3)^x

Hence eqn is just (4/3)^x = 7

Take logs with base 4/3 of both sides for x.
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