Nahmed:)
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Hiiii, i'm stuck on this question

Show that logax^10-2loga(x^3/4)=4loga(2x)

any help would be much appreciated))
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monkeyvonban
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Because they are all to the same base I am going to ignore the base and just write log(X)

Using the rule of logs that alog(x)=log(x^a) you can rewrite the equation as
log(x^10)-log((x^6)/16) [because ((x^3)/4)^2=(x^6)/16 ]

Then using the rule of logs that log(a)-log(b)=Log(a/b) we can rewrite this as

Log((x^10)/((x^6)/16)) this is the same as log((x^10)(x^-6)16) which equals log(16(x^4)) which we can use rules of logs to write as 4log(2x)
[because (2x)^4=16(x^4) ]

Not sure if this is confusingly laid out, If you don't get it I'll have another go at writing it out
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Nahmed:)
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Report Thread starter 6 years ago
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(Original post by monkeyvonban)
Because they are all to the same base I am going to ignore the base and just write log(X)

Using the rule of logs that alog(x)=log(x^a) you can rewrite the equation as
log(x^10)-log((x^6)/16) [because ((x^3)/4)^2=(x^6)/16 ]

Then using the rule of logs that log(a)-log(b)=Log(a/b) we can rewrite this as

Log((x^10)/((x^6)/16)) this is the same as log((x^10)(x^-6)16) which equals log(16(x^4)) which we can use rules of logs to write as 4log(2x)
[because (2x)^4=16(x^4) ]

Not sure if this is confusingly laid out, If you don't get it I'll have another go at writing it out
Thank you so much! i finally understand it haha... xx
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