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If Log 5 = a, and Log 2 = b, determine Log 125 + Log 8 in terms of a and b

 

I did this:

 

\(Log 125 + Log 8\)

 

\(Log 5^3 + Log 2^3\)

 

\(a^3 + b^3\)

 

OR,

 

\(3Log5 + 3Log2\)

 

\(3a + 3b\)

 

Once again, thank you..

 Nov 12, 2018

Best Answer 

 #1
avatar+24133 
+11

If Log(5) = a, and Log(2) = b, determine Log(125) + Log(8) in terms of a and b

 

\(\begin{array}{|rcll|} \hline \boxed{125 = 5^3 \\ 8 = 2^3} \\ && \log(125) + \log(8) \\ &=& \log(5^3) + \log(2^3) \quad & \quad \boxed{\log(a^b) = b\times\log(a)} \\ &=& 3\times \log(5) + 3\times \log(2) \quad & \quad \log(5) = a,~ \log(2)=b \\ &=& 3\times a + 3\times b \\ &\mathbf{=}& \mathbf{3\times (a + b)} \\ \hline \end{array}\)

 

laugh

 Nov 12, 2018
 #1
avatar+24133 
+11
Best Answer

If Log(5) = a, and Log(2) = b, determine Log(125) + Log(8) in terms of a and b

 

\(\begin{array}{|rcll|} \hline \boxed{125 = 5^3 \\ 8 = 2^3} \\ && \log(125) + \log(8) \\ &=& \log(5^3) + \log(2^3) \quad & \quad \boxed{\log(a^b) = b\times\log(a)} \\ &=& 3\times \log(5) + 3\times \log(2) \quad & \quad \log(5) = a,~ \log(2)=b \\ &=& 3\times a + 3\times b \\ &\mathbf{=}& \mathbf{3\times (a + b)} \\ \hline \end{array}\)

 

laugh

heureka Nov 12, 2018
 #2
avatar+603 
0

Hi Heureka,

I really thought I had replied, but for some reason I do not see it...I was asking yhis:

Because they asked for in terms of a and b, does it necessarilly mean that I cannot multiply the 3 with the bracket?..so 3a + 3b is incorrect?..the "a" and "b" has to be single digits?

juriemagic  Nov 14, 2018
 #3
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0

Your answer is correct! 3a + 3b is exactly the same as 3 x (a + b). He has simply factored out the 3.

 Nov 14, 2018
 #4
avatar+603 
0

thank you guest..much appreciated

juriemagic  Nov 14, 2018

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