thefatone
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#1
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#1
https://c4a3f001dcd45afe69d0ceec8300...%20Edexcel.pdf

how would one do this?

log_2 \left( 11y-3\right) - log_2 3 -log_2 y^2 =1

log_2 \left( \dfrac{11y-3}{3}\right) - log_2 y^2 =1

log_2 \left(\dfrac{y^2 \left(11y-3\right)}{3} \right)=1
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The-Spartan
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#2
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Start by using log rules to combine the logs into one single log. Then take the log away using the power (of the base) 2.
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\log_2{x}-\log_2{y}=\log_2{\frac{x}{y}}
2^{\log_2{x}} = x
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Pablo Picasso
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Log(11y-3/3 y•y) = 1
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thefatone
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#4
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(Original post by The-Spartan)
Start by using log rules to combine the logs into one single log. Then take the log away using the power (of the base) 2.
(Original post by Pablo Picasso)
Log(11y-3/3 y•y) = 1
^^ yes where did i go wrong and what do i do to correct it?
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Math12345
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Your last step is wrong. It should be 3y^2 in the denominator.

\frac{a/b}{c}=\frac{a}{bc}
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thefatone
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(Original post by Math12345)
Your last step is wrong. It should be 3y^2 in the denominator.

\frac{a/b}{c}=\frac{a}{bc}
ah oops thanks a ton
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