How do you solve the exponential equation 25^(x-1)=125^(4x)25x1=1254x?

1 Answer
Feb 25, 2017

x=-1/5x=15

Explanation:

One way to solve this exponential would be through the use of logs. But maths is about more than just parroting out formulae. It's also about trying to see connections and ways to solve equations without always resorting to the use of formula.

So instead of using logs, let's try and rewrite this equation in a way that will make it easy to solve.

We have 25^(x-1) = 125^(4x)25x1=1254x

And we know 25=5^2, 125=5^325=52,125=53

So we can rewrite 25^(x-1) = 125^(4x)25x1=1254x

as (5^2)^(x-1) = (5^3)^(4x)(52)x1=(53)4x

5^(2x-2) = 5^(12x)52x2=512x

Since we now have both exponentials raising the same number, we can set them equal to each other.

2x-2 = 12x2x2=12x

10x=-210x=2

x=-1/5x=15