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2011-02-01T00:00:01
<mdoxml version="1.0"><br></br>
<p>Without the use of calculator, find $\log_549 \times \log_7125$.</p>
<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
Several methods were used to find $\log_5 49 \times \log_7 125$.
Matthew of Queen Mary's Grammar School, Walsall changed the base to
base 10 and Labboid used a change of base to natural logarithms.
<br></br>
<br></br>
$$\eqalign{ \log_5 49 &= \ln 49 / \ln 5 \cr \log_7 125 &=
\ln 125 / \ln 7.}$$ <br></br>
<br></br>
Hence <br></br>
<br></br>
$${ {\log_5 49} \times {\log_7 125} }$$ <br></br>
<br></br>
$${ = {{\ln 49} \over {\ln 5}} \times {{\ln {125}} \over {\ln 1}}
}$$ <br></br>
<br></br>
$$ = {\ln(7^2) \times \ln(5^3)\over \ln5 \times \ln 7} $$ <br></br>
<br></br>
$$ = {2\ln 7 \times 3\ln5 \over \ln 5 \times \ln 7} $$ <br></br>
<br></br>
$$ = {2 \times 3 = 6} $$ <br></br>
<br></br>
Patrick, Saul, Hyeyoun, Marc and Arun used this identity $\log_a
b\times \log_b a = 1$ to find the solution as follows. Now $\log_5
49=2\log_5 7$ and $\log_7 125 = 3\log_7 5$ therefore <br></br>
<br></br>
$\log_5 49 \times \log_7 125 = 2\log_5 7 \times 3\log_7 5 = 6
\times \log_5 7 \times \log_7 5 = 6.$ <br></br>
<br></br>
Yatir Halevi generalized this problem to: $\log_a b^c \times \log_b
a^d$. Using the logarithms rules: <br></br>
<br></br>
$\log_a b^r= r\times\log_a b$ and $\log_b a = 1/\log_a b$ we get
the expression equal to $cd (\log_a b \times \log_b a) = cd$. So
our expression is equal to $2\times 3=6$.<br></br></mdoxml>
<mdoxml version="1.0"><br></br>
<p>If you use the fact that $\log_b a \times \log_a b=1$ you'll
need to prove this result.</p>
<br></br></mdoxml>
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Without the use of tables or calculator, find log_5 49 × log_7
125 (log_5 is log to the base 5)
Sequences, Functions and Graphs
Logarithmic functions
Sequences, Functions and Graphs
Laws of logarithms
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