Question on: WAEC Mathematics - 2013
If 2 log x (3\(\frac{3}{8}\)) = 6, find the value of x
A
\(\frac{3}{2}\)
B
\(\frac{4}{3}\)
C
\(\frac{2}{3}\)
D
\(\frac{1}{2}\)
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Correct Option: A
2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2)
\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\)
\(\log_x \frac{27}{8} = 3\)
\(\frac{27}{8} = x^3\)
\(x^3 = \frac{27}{8}\)
x = \(\sqrt{\frac{27}{8}}\)
= (\(\frac{27}{8}\))\(\frac{1}{3}\)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\)
= \(\frac{3}{2}\)
\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\)
\(\log_x \frac{27}{8} = 3\)
\(\frac{27}{8} = x^3\)
\(x^3 = \frac{27}{8}\)
x = \(\sqrt{\frac{27}{8}}\)
= (\(\frac{27}{8}\))\(\frac{1}{3}\)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\)
= \(\frac{3}{2}\)
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