Question

How do you solve `4^(3x+2)times32^(x-2)=8^(3x-4)`?

Answer 1

The soln. is `x=-3.`

Explanation:

`4^(3x+2)xx32^(x-2)=8^(3x-4)`

`rArr (2^2)^(3x+2)*(2^5)^(x-2)=(2^3)^(3x-4)`

`rArr 2^(6x+4)*2^(5x-10)=2^(9x-12).........[(a^m)^n=a^(mn)]`

`rArr 2^(6x+4+5x-10)=2^(9x-12)...............[a^m*a^n=a^(m+n)]`

`rArr 2^(11x-6)=2^(9x-12)..............[a^m=a^n rArr m=n]`

`rArr 11x-6=9x-12`

`rArr 11x-9x=6-12`

`rArr 2x=-6`

`rArr x=-3`

This root satisfy the eqn.

Hence, the soln. is `x=-3.`

Answer 2

`x = - 3`

Explanation:

We have: `4^(3 x + 2) times 32^(x - 2) = 8^(3 x - 4)`

Let's express the numbers in terms of `2`:

`=> (2^(2))^(3 x + 2) times (2^(5))^(x - 2) = (2^(3))^(3 x - 4)`

Using the laws of exponents:

`=> 2^(6 x + 4) times 2^(5 x - 10) = 2^(9 x - 12)`

`=> 2^(6 x + 4 + 5 x - 10) = 2^(9 x - 12)`

`=> 2^(11 x - 6) = 2^(9 x - 12)`

`=> 11 x - 6 = 9 x - 12`

`=> 2 x = - 6`

`=> x = - 3`

Therefore, the solution to the equation is `x = - 3`.