If ${log}_{b} a = \frac{1}{x} and {log}_{a} \sqrt b = 3 x^{2}$ $log_ba=1/x and log_a √b =3x^2$ , show that x =1/6?

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If ${log}_{b} a = \frac{1}{x} and {log}_{a} \sqrt b = 3 x^{2}$ $log_ba=1/x and log_a √b =3x^2$ , show that x =1/6?

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Answer 1 · 2018-06-03T18:21:08

$see explanation$ $"see explanation"$

Explanation:

$using the law of logarithms$ $"using the "color(blue)"law of logarithms"$

$∙ x {log}_{b} x = n \Leftrightarrow x = b^{n}$ $•color(white)(x)log_b x=nhArrx=b^n$

${log}_{b} a = \frac{1}{x} \Rightarrow a = b^{\frac{1}{x}}$ $log_b a=1/xrArra=b^(1/x)$

${log}_{a} b^{\frac{1}{2}} = 3 x^{2} \Rightarrow b^{\frac{1}{2}} = a^{3 x^{2}}$ $log_a b^(1/2)=3x^2rArrb^(1/2)=a^(3x^2)$

$substitute a = b^{\frac{1}{x}} into a^{3 x^{2}}$ $"substitute "a=b^(1/x)" into "a^(3x^2)$

$b^{\frac{1}{2}} = {(b^{\frac{1}{x}})}^{3 x^{2}} = b^{3 x}$ $b^(1/2)=(b^(1/x))^(3x^2)=b^(3x)$

$we have b^{\frac{1}{2}} = b^{3 x}$ $"we have "b^(1/2)=b^(3x)$

$3 x = \frac{1}{2}$ $3x=1/2$

$divide both sides by 3$ $"divide both sides by 3"$

$x = \frac{\frac{1}{2}}{3} = \frac{1}{6}$ $x=(1/2)/3=1/6$

Answer 2 · 2018-06-03T18:21:31

$x = \frac{1}{6}$ $x=1/6$

Explanation:

${log}_{b} a = \frac{1}{x}$ $log_b a=1/x$

So

$x {log}_{b} a = 1$ $xlog_ba=1$

$x = \frac{1}{{log}_{b} a}$ $x=1/log_b a$

$x = {log}_{a} b$ $x=log_ab$

Now

$\frac{1}{2} {log}_{a} b = 3 x^{2}$ $1/2log_a b=3x^2" "$ (substituting from ${log}_{a} b = x$ $log_a b=x$ )

So

$\frac{x}{2} = 3 x^{2}$ $x/2=3x^2$

$x = 6 x^{2}$ $x=6x^2$

$6 x = 1$ $6x=1$

$x = \frac{1}{6}$ $x=1/6$ with $x \neq 0$ $x !=0$

Hence proved.

Answer 3 · 2018-06-03T18:38:56

We will start the proof

Explanation:

We have
$\frac{ln (a)}{ln (b)} = \frac{1}{x}$ $ln(a)/ln(b)=1/x$
and

$\frac{1}{2} \cdot \frac{ln (b)}{ln (a)} = 3 x^{2}$ $1/2*ln(b)/ln(a)=3x^2$
so

$\frac{ln (a)}{ln (b)} = \frac{1}{6 x^{2}}$ $ln(a)/ln(b)=1/(6x^2)$

using the first equation we get
$\frac{1}{x} = \frac{1}{6 x^{2}}$ $1/x=1/(6x^2)$

multiplying by $x^{2}$ $x^2$ we get

$x = \frac{1}{6}$ $x=1/6$ for $x \neq 0$ $xne 0$

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If ${log}_{b} a = \frac{1}{x} and {log}_{a} \sqrt b = 3 x^{2}$ $log_ba=1/x and log_a √b =3x^2$ , show that x =1/6?

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Explanation:

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... and beyond

If log_ba=1/x and log_a √b =3x^2logba=1xandloga√b=3x2log_ba=1/x and log_a √b =3x^2, show that x =1/6?

3 Answers

Explanation:

Explanation:

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If ${log}_{b} a = \frac{1}{x} and {log}_{a} \sqrt b = 3 x^{2}$ $log_ba=1/x and log_a √b =3x^2$ , show that x =1/6?