How do you solve ${log}_{2} (x) + {log}_{6} (x) = 3$ $log_2 (x) + log_6 (x) = 3$ ?

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How do you solve ${log}_{2} (x) + {log}_{6} (x) = 3$ $log_2 (x) + log_6 (x) = 3$ ?

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Answer 1 · 2015-09-02T15:56:41

$x = 2^{\frac{3 ({log}_{2} 3 + 1)}{{log}_{2} 3 + 2}}$ $x = 2^((3(log_2 3+1))/(log_2 3 + 2))$

Explanation:

${log}_{a} b = \frac{{log}_{c} b}{{log}_{c} a}$ $log_a b = log_c b/log_c a$
$log (a b) = log a + log b$ $log (ab) = log a + log b$
${log}_{a} a = 1$ $log_a a = 1$
${log}_{a} b = c \Leftrightarrow a^{c} = b$ $log_a b = c <=> a^c = b$

${log}_{2} x + {log}_{6} x = 3$ $log_2 x + log_6 x = 3$
${log}_{2} x + \frac{{log}_{2} x}{{log}_{2} 6} = 3$ $log_2 x + log_2 x/log_2 6 = 3$
${log}_{2} x (1 + \frac{1}{{log}_{2} 6}) = 3$ $log_2 x(1+1/log_2 6) = 3$
${log}_{2} x = \frac{3 {log}_{2} 6}{{log}_{2} 6 + 1}$ $log_2 x = (3log_2 6)/(log_2 6 + 1)$
$x = 2^{\frac{3 {log}_{2} 6}{{log}_{2} 6 + 1}}$ $x = 2^((3log_2 6)/(log_2 6 + 1))$
$x = 2^{\frac{3 ({log}_{2} 3 + 1)}{{log}_{2} 3 + 2}}$ $x = 2^((3(log_2 3+1))/(log_2 3 + 2))$

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How do you solve ${log}_{2} (x) + {log}_{6} (x) = 3$ $log_2 (x) + log_6 (x) = 3$ ?

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Explanation:

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How do you solve log2(x)+log6(x)=3log_2 (x) + log_6 (x) = 3?

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