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

1 Answer
May 30, 2015

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