Question

Q7. Find the value of the expression:
⠀⠀
$\sf{\dfrac{2}{\log_4(2000)^6}+\dfrac{3}{\log_5(2000)^6}}$​

Answer 1

★ Concept :-

Here the concept of Logarithmic Properties has been used. We see that we are given two fractions whose resultant we need to find out. Firstly we can simplify the denominators of both the equation. Then we can apply logarithmic properties there. After that we have apply some Logarithmic values and then by arithmetic operations, we have to solve the qúestion.

Let's do it !!

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★ Formula Used :-

$\;\;\boxed{\sf{\pink{\log_{y}x^{a}\;=\;a\log_{y}x}}}$

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★ Solution :-

Given,

$\;\;\rm{\rightarrow\;\;\green{\dfrac{2}{\log_{4}(2000)^{6}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}}$

Now let's start solving this. We will solve each fraction seperately then we will add them.

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}(2000)^{6}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

We know that,

• 2000 = 2 × 1000 = 2 × 10³

By applying this in the initial expression,

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}(2\:\times\:1000)^{6}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}(2\:\times\:10^{3})^{6}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

We know that,

• (ab)ⁿ = aⁿ × bⁿ

• Here a = 2

• Here b = 10³

• Here n = 6

By applying this in the first fraction, we get

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}[(2)^{6}\:\times\:(10^{3})^{6}]}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

We know that,

$\;\qquad\quad\odot\;\;\bf{(a^{n})^{m}\;=\;a^{nm}}$

• Here a = 10

• Here n = 3

• Here m = 6

By applying this here, we get

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}[(2)^{6}\:\times\:(10)^{3\:\times\:6}]}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}[(2)^{6}\:\times\:(10)^{18}]}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

• log(ab) = log a + log b

This can be written as,

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{\log_{4}[(2)^{6}]\:+\:\log_{4}[(10)^{18}]}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

We know that,

$\;\qquad\quad\odot\;\;\bf{\log_{y}x^{a}\;=\;a\log_{y}x}$

On applying this formula here, we get

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\log_{4}(2)\:+\:18\log_{4}(10)}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

Now here 6 can be taken common from first fraction.

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2)\:+\:3\log_{4}(10)\}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2)\:+\:\log_{4}(10)^{3}\}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2)\:+\:\log_{4}(1000)\}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

Now again using the property of log(ab) = log a + log b

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2\:\times\:1000)\}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{\log_{5}(2000)^{6}}}$

Now let's start solving the second fraction by similar process.

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{\log_{5}(2\:\times\:10^{3})^{6}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{\log_{5}[(2)^{6}\:\times\:(10^{3})^{6}]}}$

(since log(ab) = log a + log b)

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{6\log_{5}(2)\:+\:18\log_{5}(10)}}$

Taking 6 in common, we get

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{6\{\log_{5}(2)\:+\:\log_{5}(1000)\}}}$

Again using the property of log(ab) = log a + log b

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{6\{\log_{5}(2\:\times\:1000)\}}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2}{6\{\log_{4}(2000)\}}\;+\;\dfrac{3}{6\{\log_{5}(2000)\}}}$

We know that,

$\;\qquad\quad\odot\;\;\bf{\dfrac{1}{\log_{y}x}\;=\;\log_{x}y}$

On applying this in both fractions we get,

$\;\;\tt{\Longrightarrow\;\;\dfrac{2\:\times\:\{\log_{2000}(4)\}}{6}\;+\;\dfrac{3\:\times\:\{\log_{2000}(5)\}}{6}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{2\log_{2000}(4)}{6}\;+\;\dfrac{3\log_{2000}(5)}{6}}$

Now again using the property,

$\;\qquad\quad\odot\;\;\bf{\log_{y}x^{a}\;=\;a\log_{y}x}$

we get,

$\;\;\tt{\Longrightarrow\;\;\dfrac{\log_{2000}(4)^{2}}{6}\;+\;\dfrac{\log_{2000}(5)^{3}}{6}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{\log_{2000}(16)}{6}\;+\;\dfrac{\log_{2000}(125)}{6}}$

Since denominators are same, so this can be written as,

$\;\;\tt{\Longrightarrow\;\;\dfrac{\log_{2000}(16)\;+\;\log_{2000}(125)}{6}}$

Here again we shall apply the property of log a + log b = log(ab)

So,

$\;\;\tt{\Longrightarrow\;\;\dfrac{\log_{2000}(16\:\times\:125)}{6}}$

$\;\;\tt{\Longrightarrow\;\;\dfrac{\log_{2000}(2000)}{6}}$

We know that,

$\;\qquad\quad\odot\;\;\bf{\log_{x}x\;=\;1}$

On applying this, we get

$\;\;\bf{\Longrightarrow\;\;\red{\dfrac{1}{6}}}$

This is the required answer.

$\;\;\underline{\boxed{\tt{Required\;\:Answer\;\;=\;\bf{\purple{\dfrac{1}{6}}}}}}$

Answer 2

It is the question of resolution and reduction...