Question

Please solve q 5?

Answer 1

See Below.

Explanation:

We Have,

`color(white)(xxx)a^(2x - 3) * b^(2x) = a^(6-x) * b^(5x)`

`rArr a^(2x - 3)/a^(6-x) = b^(5x)/b^(2x)`

[Just Transpose the `a` and `b` on their respective sides.]

`rArr a^((2x - 3)-(6 - x)) = b^(5x - 2x)` [As `a^(m-n) = a^m/a^n`]

`rArr a^(2x - 3 - 6 +x) = b^(3x)`

`rArr a^(3x - 9) = b^(3x)`

`rArr (a^(x - 3))^3 = (b^x)^3` [As, `(x^m)^n = x^(mn)`]

`rArr a^(x - 3) = b^x`

`rArr a^x/a^3 = b^x` [As `a^(m-n) = a^m/a^n`]

`rArr a^x/b^x = a^3` [Transposing again]

`rArr (a/b)^x = a^3` [As `(a/b)^m = a^m/b^m`]

`rArr log (a/b)^x = log a^3` [Taking `log` at both sides]

`rArr x log (a/b) = 3 log a` [As `log a^x = x log a`]

`rArr 3 log a = x log (a/b)`

Hence Proved.