Question

64^x + 64^2x + 64^3x = 84 ..x=?​

Answer 1

Required Answer:-

Given:

• $\tt {64}^{x} + {64}^{2x} + {64}^{3x} = 84$

To Find:

• The value of x.

Solution:

Given equation –

$\tt \rightarrow {64}^{x} + {64}^{2x} + {64}^{3x} = 84$

This can be written as –

$\tt \rightarrow {( {2}^{6} )}^{x} + {({2}^{6})}^{2x} + {( {2}^{6} )}^{3x} = 84$

As we know that,

$\tt \maltese \: ( {x}^{a})^{b} = {x}^{ab}$

Therefore,

$\tt \rightarrow {2}^{6x} + {2}^{12x} + {2}^{18x} = 84$

Let us assume that,

$\tt \rightarrow y = {2}^{6x}$

So, the equation becomes,

$\tt \rightarrow y + {y}^{2} + {y}^{3} = 84$

$\tt \rightarrow {y}^{3} + {y}^{2} + y - 84 = 0$

Now, solve the equation by factorisation.

$\tt \rightarrow {y}^{3} + (5 - 4) {y}^{2} + (21 - 20)y - 84 = 0$

By splitting the terms, we get,

$\tt \rightarrow {y}^{3} +5 {y}^{2} - 4{y}^{2} + 21y - 20y - 84 = 0$

$\tt \rightarrow {y}^{3} - 4{y}^{2} + 5{y}^{2} - 20y + 21y - 84 = 0$

$\tt \rightarrow {y}^{2}(y - 4) + 5y(y-4) + 21(y - 4) = 0$

$\tt \rightarrow ( {y}^{2} + 5y + 21)(y - 4)= 0$

Therefore,

$\tt \rightarrow ( {y}^{2} + 5y + 21) = 0 \: or \: (y - 4)= 0$

In case of y² + 5y + 21 = 0, no real roots exists.

So,

$\tt \rightarrow (y - 4)= 0$

$\tt \rightarrow y = 4$

Now, substitute back,

$\tt \rightarrow y = {2}^{6x}$

$\tt \rightarrow {2}^{6x} = 4$

$\tt \rightarrow {2}^{6x} = {2}^{2}$

Comparing base, we get,

$\tt \rightarrow 6x = 2$

$\tt \rightarrow x = \dfrac{2}{6}$

$\tt \rightarrow x = {}^{1}/_{3}$

✠ So, the value of x is 1/3.

Answer:

• x = 1/3