Question

(1) log2 x + logs 8x + log32 x = 15​

Answer 1

Appropriate Question : -

Solve for x

$\rm :\longmapsto\: log_{2}(x) + log_{8}(x) + log_{32}(x) = 15$

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

Given Logarithmic equation is

$\rm :\longmapsto\: log_{2}(x) + log_{8}(x) + log_{32}(x) = 15$

can be rewritten as

$\rm :\longmapsto\: log_{2}(x) + log_{ {2}^{3} }(x) + log_{ {2}^{5} }(x) = 15$

We know,

$\rm :\longmapsto\:\boxed{\tt{ log_{ {x}^{a} }( {y}^{b} ) = \frac{b}{a} log_{x}(y) \: }}$

So, using this identity, we get

$\rm :\longmapsto\: log_{2}(x) + \dfrac{1}{3}log_{2}(x) + \dfrac{1}{5}log_{2}(x) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(1 + \dfrac{1}{3} + \dfrac{1}{5} \bigg) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(\dfrac{15 + 5 + 3}{15} \bigg) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(\dfrac{23}{15} \bigg) = 15$

$\rm :\longmapsto\:log_{2}x = 15 \times \dfrac{15}{23}$

$\rm :\longmapsto\:log_{2}x = \dfrac{225}{23}$

We know,

$\boxed{\tt{ log_{a}(b) = c \: \rm\implies \: \: b = {a}^{c} \: }}$

So, using this, we get

$\bf\implies \:x = {\bigg(2\bigg) }^{\dfrac{225}{23} }$

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Additional Information

$\boxed{\tt{ log_{x}(x) = 1}}$

$\boxed{\tt{ log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} }}$

$\boxed{\tt{ {a}^{ log_{a}(x) } = x}}$

$\boxed{\tt{ {a}^{ ylog_{a}(x) } = {x}^{y} }}$

$\boxed{\tt{ log( {x}^{y} ) = y \: logx \: }}$

Answer 2

Answer:

Appropriate Question : -

Solve for x

$\rm :\longmapsto\: log_{2}(x) + log_{8}(x) + log_{32}(x) = 15$

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

Given Logarithmic equation is

$\rm :\longmapsto\: log_{2}(x) + log_{8}(x) + log_{32}(x) = 15$

can be rewritten as

$\rm :\longmapsto\: log_{2}(x) + log_{ {2}^{3} }(x) + log_{ {2}^{5} }(x) = 15$

We know,

$\rm :\longmapsto\:\boxed{\tt{ log_{ {x}^{a} }( {y}^{b} ) = \frac{b}{a} log_{x}(y) \: }}$

So, using this identity, we get

$\rm :\longmapsto\: log_{2}(x) + \dfrac{1}{3}log_{2}(x) + \dfrac{1}{5}log_{2}(x) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(1 + \dfrac{1}{3} + \dfrac{1}{5} \bigg) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(\dfrac{15 + 5 + 3}{15} \bigg) = 15$

$\rm :\longmapsto\:log_{2}(x)\bigg(\dfrac{23}{15} \bigg) = 15$

$\rm :\longmapsto\:log_{2}x = 15 \times \dfrac{15}{23}$

$\rm :\longmapsto\:log_{2}x = \dfrac{225}{23}$

We know,

$\boxed{\tt{ log_{a}(b) = c \: \rm\implies \: \: b = {a}^{c} \: }}$

So, using this, we get

$\bf\implies \:x = {\bigg(2\bigg) }^{\dfrac{225}{23} }$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\boxed{\tt{ log_{x}(x) = 1}}$

$\boxed{\tt{ log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} }}$

$\boxed{\tt{ {a}^{ log_{a}(x) } = x}}$

$\boxed{\tt{ {a}^{ ylog_{a}(x) } = {x}^{y} }}$

$\boxed{\tt{ log( {x}^{y} ) = y \: logx \: }}$