Question

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Solve the logarithmic inequality for real values of x

$log_{0.04}(x - 1) \geqslant log_{0.2}(x - 1)$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic inequality is

$\rm :\longmapsto\: log_{0.04}(x - 1) \geqslant log_{0.2}(x - 1)$

can be rewritten as

$\rm :\longmapsto\: log_{ {(0.2)}^{2} }(x - 1) \geqslant log_{0.2}(x - 1)$

We know,

$\red{ \boxed{ \sf{ \: log_{ {x}^{y} }( {w}^{z} ) = \frac{z}{y} log_{x}(w) }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{2}log_{0.2}(x - 1) \geqslant log_{0.2}(x - 1)$

$\rm :\longmapsto\:log_{0.2}(x - 1) \geqslant 2log_{0.2}(x - 1)$

$\rm :\longmapsto\:log_{0.2}(x - 1) - 2log_{0.2}(x - 1) \geqslant 0$

$\rm :\longmapsto\: - \: log_{0.2}(x - 1) \geqslant 0$

$\rm :\longmapsto\: \: log_{0.2}(x - 1) \leqslant 0$

We know,

$\red{ \boxed{ \sf{ 0 < a < 1 \: and \: \: log_{a}(b) \leqslant c \: \implies \: b \geqslant {a}^{c}}}}$

So, using this, we get

$\rm :\longmapsto\:x - 1 \geqslant {(0.2)}^{0}$

$\rm :\longmapsto\:x - 1 \geqslant 1$

$\rm :\longmapsto\:x \geqslant 1 + 1$

$\rm :\longmapsto\:x \geqslant 2$

$\bf\implies \:x \: \in \: [2, \infty )$

Additional Information :-

$\red{ \boxed{ \sf{ \: log_{a}(a) = 1}}}$

$\red{ \boxed{ \sf{ \: log_{ {a}^{p} }( {a}^{q} ) = \frac{q}{p} }}}$

$\red{ \boxed{ \sf{ \: log_{x}(y) = \frac{1}{ log_{y}(x) }}}}$

$\red{ \boxed{ \sf{ \: {x}^{ log_{x}(y) } = y}}}$

$\red{ \boxed{ \sf{ \: {x}^{ zlog_{x}(y) } = {y}^{z} }}}$