How do you solve ${log}_{b} (x^{2} - 2) + 2 {log}_{b} 6 = {log}_{b} 6 x$ $log_b(x^2 -2) + 2log_b 6 = log_b6x$ ?

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Answer 1 · 2016-05-21T10:22:47

$x = {\frac{3}{2}, \frac{4}{3}}$ $x={3/2,4/3}$

${log}_{b} (x^{2} - 2) + 2 {log}_{b} 6 = {log}_{b} 6 x$ $log_b(x^2-2)+2log_b6=log_b6x$

${log}_{b} (x^{2} - 2) + {log}_{b} 6^{2} = {log}_{b} 6 x$ $log_b(x^2-2)+log_b6^2=log_b6x$

${log}_{b} (x^{2} - 2) \cdot 6^{2} = {log}_{b} 6 x$ $log_b(x^2-2)*6^2=log_b6x$

${log}_{b} (x^{2} - 2) \cdot 6^{2} - {log}_{b} 6 x = 0$ $log_b(x^2-2)*6^2-log_b6x=0$

${log}_{b} (\frac{(x^{2} - 2) \cdot 6^{2}}{6 x}) = 0$ $log_b(((x^2-2)*6^2)/(6x))=0$

${log}_{b} (\frac{6 x^{2} - 12}{x}) = 0$ $log_b((6x^2-12)/x)=0$

$\frac{6 x^{2} - 12}{x} = b^{0}$ $(6x^2-12)/x=b^0$

$b^{0} = 1$ $b^0=1$

$\frac{6 x^{2} - 12}{x} = 1$ $(6x^2-12)/x=1$

$6 x^{2} - 12 = x 6 x^{2} - x - 12 = 0$ $6x^2-12=x" "6x^2-x-12=0$

$(2 x - 3) (3 x + 4) = 0$ $(2x-3)(3x+4)=0$

$2 x - 3 = 02 x = 3 x = \frac{3}{2}$ $2x-3=0" "2x=3" " x=3/2$

$3 x + 4 = 03 x = 4 x = \frac{4}{3}$ $3x+4=0" "3x=4" "x=4/3$

$x = {\frac{3}{2}, \frac{4}{3}}$ $x={3/2,4/3}$

How do you solve log_b(x^2 -2) + 2log_b 6 = log_b6xlogb(x2−2)+2logb6=logb6xlog_b(x^2 -2) + 2log_b 6 = log_b6x?