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If loga(s) = 5 and loga(t) = 10, then loga(s/t2) = ?

 Jun 27, 2018
 #1
avatar+102905 
+1

loga s  = 5    loga t   =10

 

In exponential form, we have

 

a^5  = s       and  a^10  =  t

 

So

 

loga (s / t^2 )  =

 

loga [a^5 / (a^10)^2 ]  =

 

loga [ a^5 / a^20 ]  =

 

loga  a^(-15)        and by a log property we can write

 

-15 * loga a  =

 

 ( Note  loga a   =  1  )

 

-15 * 1  =

 

-15

 

cool cool cool

 Jun 27, 2018
edited by CPhill  Jun 27, 2018
 #2
avatar+23040 
0

If loga(s) = 5 and loga(t) = 10, then loga(s/t2) = ?

 

\(\begin{array}{|rcll|} \hline \log_a(s) &=& 5 \\ \log_a(t) &=& 10 \\\\ \log_a(s) - 2\cdot \log_a(t) &=& 5 - 2\cdot 10 \\ \log_a(s) - 2\cdot \log_a(t) &=& -15 \\ \log_a(s) -\log_a(t^2) &=& -15 \\\\ \mathbf{\log_a\left(\dfrac{s}{t^2} \right)} & \mathbf{=} & \mathbf{-15} \\ \hline \end{array}\)

 

laugh

 Jun 28, 2018

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