Question

How do you solve `64^(x-1)/16^(2x+2)=2^(x-2)`?

Answer 1

`x = 4`

Explanation:

`64^(x-1)/16^(2x+2)=2^(x-2)`

Well. We know that `64= 2^6` and `16 = 2^4`. Also we know that

`(a^b)^c = a^(bc)`

so

`64^(x-1)/16^(2x+2)=2^(x-2) equiv (2^6)^(x-1)/(2^4)^(2x+2)=2^(x-2)`

and also

`64^(x-1)/16^(2x+2)=2^(x-2) equiv 2^(6(x-1))/2^(4(2x+2))=2^(x-2)`

Now, applying the `log_2` transformation to both sides and keeping in mind that

`log_a a^b = b`

`6(x-1)-4(2x+2)=x-2`. Now, solving for `x`

`x =4`

Answer 2

`x= -4`

Explanation:

All the bases are powers of 2 - change them all to a base of 2.
Then we can use laws of indices.

`(color(red)(64)^(x-1))/(color(blue)(16)^(2x+2)) = 2^(x-2)`

`((color(red)(2^6))^(x-1))/((color(blue)(2^4))^(2x+2)) = 2^(x-2) color(white)(xxxxxxxxxx)` Multiply the indices.

`2^(color(magenta)(6x-6))/(2^(color(magenta)(8x+8))) = 2^(x-2)color(white)(xxxxxxxxxxxx)` Subtract the indices

`2^(color(magenta)(-2x-14)) = 2^(x-2 )color(white)(xxxxxxxxxxxx)`Equate the indices.

`-2x-14 = x-2 color(white)(xxxxxxxxxx)` Solve for x

`-14+2 = x+2x`

`-12 = 3x`

`x = -4`