How do you solve 5^x = 25^(x-1) 5x=25x−1?

1 Answer
Elvan Burcu K.
May 30, 2015

First thing we should know is;
25^b=(5^2)^b=5^(2b)25b=(52)b=52b This is a basic rule to be memorised.
So our question is;
5^x=25^(x-1) 5x=25x−1
We can write it as ;
5^x=(5^2)^(x-1) => 5^x= 5^(2*(x-1)) => 5^x=5^(2x-2)5x=(52)x−1⇒5x=52⋅(x−1)⇒5x=52x−2 Since bases (5 values) are same. We can write;
x=2x-2 => ul (x=2)

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