+1124 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..
+26393 If Log(5) = a, and Log(2) = b, determine Log(125) + Log(8) in terms of a and b
\(\begin{array}{|rcll|} \hline \boxed{125 = 5^3 \\ 8 = 2^3} \\ && \log(125) + \log(8) \\ &=& \log(5^3) + \log(2^3) \quad & \quad \boxed{\log(a^b) = b\times\log(a)} \\ &=& 3\times \log(5) + 3\times \log(2) \quad & \quad \log(5) = a,~ \log(2)=b \\ &=& 3\times a + 3\times b \\ &\mathbf{=}& \mathbf{3\times (a + b)} \\ \hline \end{array}\)
+26393 If Log(5) = a, and Log(2) = b, determine Log(125) + Log(8) in terms of a and b
\(\begin{array}{|rcll|} \hline \boxed{125 = 5^3 \\ 8 = 2^3} \\ && \log(125) + \log(8) \\ &=& \log(5^3) + \log(2^3) \quad & \quad \boxed{\log(a^b) = b\times\log(a)} \\ &=& 3\times \log(5) + 3\times \log(2) \quad & \quad \log(5) = a,~ \log(2)=b \\ &=& 3\times a + 3\times b \\ &\mathbf{=}& \mathbf{3\times (a + b)} \\ \hline \end{array}\)
+1124 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?
