How do you solve ${log}_{11} 2 + 2 {log}_{11} x = {log}_{11} 32$ $log_11 2+2log_11x=log_11 32$ ?

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How do you solve ${log}_{11} 2 + 2 {log}_{11} x = {log}_{11} 32$ $log_11 2+2log_11x=log_11 32$ ?

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Answer 1 · 2016-10-27T20:43:15

$x = 4$ $x=4$

Explanation:

${log}_{11} 2 + 2 {log}_{11} x = {log}_{11} 32$ $log_11 2 +2 log_11 x = log _ 11 32$

${log}_{11} 2 + 2 {log}_{11} x - {log}_{11} 32 = 0$ $log_11 2 +2 log_11 x - log _ 11 32=0$

${log}_{11} 2 + {log}_{11} x^{2} - {log}_{11} 32 = 0$ $log_11 2 + log_11 x^2 - log _ 11 32=0$ ---> Use property ${log}_{b} x^{n} = n {log}_{b} x$ $log_bx^n=nlog_bx$

${log}_{11} (\frac{2 x^{2}}{32}) = 0$ $log_11 ((2x^2)/32)=0$ ---> Use properties ${log}_{b} x + {log}_{b} y = {log}_{b} (x y)$ $log_bx+log_by=log_b(xy)$ and #log_bx-log_by=log_b(x/y)

$11^{0} = \frac{x^{2}}{16}$ $11^0=(x^2)/16$

$1 = \frac{x^{2}}{16}$ $1=(x^2)/16$

$x^{2} = 16$ $x^2=16$

$x = \pm \sqrt{16}$ $x=+- sqrt 16$

$x = \pm 4$ $x=+-4$

$:.x=4$

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How do you solve ${log}_{11} 2 + 2 {log}_{11} x = {log}_{11} 32$ $log_11 2+2log_11x=log_11 32$ ?

1 Answer

Explanation:

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How do you solve log_11 2+2log_11x=log_11 32log112+2log11x=log1132log_11 2+2log_11x=log_11 32?

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How do you solve ${log}_{11} 2 + 2 {log}_{11} x = {log}_{11} 32$ $log_11 2+2log_11x=log_11 32$ ?