Question

If `(1/125)^(a^2 + 4ab) = (3sqrt625)^(3a^2 - 10ab)` and if `a and b` do not equal 0.. `a/b = ........` ?

Answer 1

See below.

Explanation:

Applying `log` to both sides

`(a^2 + 4 a b) Log_e(1/125) - (3 a^2 - 10 a b) Log_e(3 xx25)=0`

Now solving for `a` we get at

`{(a = 0),(a =(2 b Log_e(3^5 5^4))/(3 Log_e(3 xx 5^3)) ):}`

and finally

`a/b = (2 log_e(3^5 5^4))/(3 log_e(3 xx 5^3)) approx 1.34199`

Answer 2

`a/b = 4/21`

Explanation:

Using Indices

`(1/125)^(a^2 + 4ab) = (root(3) 625)^(3a^2 - 10ab)`

Recall that `rArr 1/a = a^-1`

`(125^-1)^(a^2 + 4ab) = (625^(1/3))^(3a^2 - 10ab)`

`(5^(3(-1)))^(a^2 + 4ab) = (5^(4(1/3)))^(3a^2 - 10ab)`

`5^(-3(a^2 + 4ab)) = 5^(4/3(3a^2 - 10ab))`

`cancel5^(-3(a^2 + 4ab)) = cancel5^(4/3(3a^2 - 10ab))`

`-3(a^2 + 4ab) = 4/3(3a^2 - 10ab)`

`-3a^2 - 12ab = (12a^2)/3 - (40ab)/3`

`-3a^2 - 12ab = (12a^2 - 40ab)/3`

Cross Multiply

`3(-3a^2 - 12ab) = 12a^2 - 40ab`

`-9a^2 - 36ab = 12a^2 - 40ab`

Collect Like Terms

`-9a^2 - 12a^2 = - 40ab + 36ab`

`-21a^2 = - 4ab`

`cancel-21a^2 = cancel- 4ab`

`21a^2 = 4ab`

Since we are looking for `color(white)(x) a/b`

Divide both sides by `ab`

`(21cancel(a^2)^1)/(cancelab) = (4cancel(ab))/(cancel(ab))`

`rArr (21a)/b = 4`

Divide both sides by `21`

`rArr ((21a)/b)/21 = 4/21`

`rArr ((cancel21a)/b) xx 1/cancel21 = 4/21`

`rArr a/b = 4/21 -> Answer`

Hence Option. A is the final Answer..