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AS Maths help! Logarithms

charlyb

Find the exact value of x for which
log2(2x)=log2(5x+4) -3

How is this done?

Reply 1

the bear

Original post by charlyb

Find the exact value of x for which
log2(2x)=log2(5x+4) -3

How is this done?

log₂X - log₂( 5X + 4 ) = -3

now simplify the LHS into one log

then rearrange according to the rule

log_pq = r <=====> q = p^r

Reply 2

charlyb

Thank you !!

I have simplified it to log2(2x/5x+4) = -3 but now I am stuck because I thought a negative number was not able to be turned into a log ?

Reply 3

charlyb

Original post by the bear

log₂X - log₂( 5X + 4 ) = -3

now simplify the LHS into one log

then rearrange according to the rule

log_pq = r <=====> q = p^r

Thank you !!

I have simplified it to log2(2x/5x+4) = -3 but now I am stuck because I thought a negative number was not able to be turned into a log ?

Reply 4

the bear

Original post by charlyb

Thank you !!

I have simplified it to log2(2x/5x+4) = -3 but now I am stuck because I thought a negative number was not able to be turned into a log ?

have a go at using the {sub}2{/sub} method for writing the bases ( use [] instead of {} ) !

negative numbers can make perfectly good logarithms with a little patience & understanding. you cannot find the log of a negative number, neither can you use a negative number as the base.

AS Maths help! Logarithms

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