Question

If 9 ^ x + 6 ^ x = 2.4 ^ x then the value of (x ^ 3 + 64)/5​

Answer 1

$\displaystyle \sf The\: value \: of \: \: \: \frac{ {x}^{3} + 64}{5} = \bf \: \frac{64}{5}$

Given :

$\displaystyle \sf {9}^{x} + {6}^{x} = 2. {4}^{x}$

To find :

$\displaystyle \sf The\: value \: of \: \: \: \frac{ {x}^{3} + 64}{5}$

Solution :

Step 1 of 3 :

Write down the given equation

Here the given equation is

$\displaystyle \sf {9}^{x} + {6}^{x} = 2. {4}^{x}$

Step 2 of 3 :

Find the value of x

$\displaystyle \sf {9}^{x} + {6}^{x} = 2. {4}^{x}$

$\displaystyle \sf \implies \frac{{9}^{x}}{{4}^{x}} + \frac{{6}^{x}}{{4}^{x}} = 2$

$\displaystyle \sf \implies {\bigg( \frac{9}{4} \bigg)}^{x} + {\bigg( \frac{6}{4} \bigg)}^{x} = 2$

$\displaystyle \sf \implies {\bigg( \frac{3}{2} \bigg)}^{2x} + {\bigg( \frac{3}{2} \bigg)}^{x} - 2 = 0$

$\displaystyle \sf Let \: \: {\bigg( \frac{3}{2} \bigg)}^{x} = a$

Then above becomes

$\displaystyle \sf {a}^{2} + a - 2 = 0$

$\displaystyle \sf{ \implies }{a}^{2} + (2 - 1)a - 2 = 0$

$\displaystyle \sf{ \implies }{a}^{2} + 2a -a - 2 = 0$

$\displaystyle \sf{ \implies }a(a + 2) - 1(a + 2) = 0$

$\displaystyle \sf{ \implies }(a + 2) (a - 1) = 0$

$\displaystyle \sf{ \implies } Either \: \: \: (a + 2) = 0 \: \: or \: \: (a - 1) = 0$

Now,

$\displaystyle \sf a + 2 = 0 \: \: gives$

$\displaystyle \sf a = - 2$

$\displaystyle \sf {\bigg( \frac{3}{2} \bigg)}^{x} = - 2$

From above value of x can not be determined

Again ,

$\displaystyle \sf a - 1 = 0 \: \: gives$

$\displaystyle \sf{ \implies }a = 1$

$\displaystyle \sf{ \implies } {\bigg( \frac{3}{2} \bigg)}^{x} = 1$

$\displaystyle \sf{ \implies } {\bigg( \frac{3}{2} \bigg)}^{x} = {\bigg( \frac{3}{2} \bigg)}^{0}$

$\displaystyle \sf{ \implies }x = 0$

So the value of x satisfying the given equation is 0

Step 3 of 3 :

$\displaystyle \sf Find\: value \: of \: \: \: \frac{ {x}^{3} + 64}{5}$

$\displaystyle \sf \frac{ {x}^{3} + 64}{5}$

$\displaystyle \sf = \frac{ {0}^{3} + 64}{5}$

$\displaystyle \sf = \frac{ 0 + 64}{5}$

$\displaystyle \sf = \frac{ 64}{5}$

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