+309 Answer: \(3\)
Solution:
\(5^b\cdot5\) (essentially what you wrote except using multiplication) can be turned into \(5^{b+1}\). \(25^{b-1}\) can be turned into \(5^{2(b-1)}\), or \(5^{2b-2}\). Now you have to set these two equal to each other:
\(5^{b+1}\)= \(5^{2b-2}\)
You can get rid of the base, because they're the same.
\(b+1=2b-2\)
Subtracting \(b\) from both sides and adding 2 gives \(b=3\)
+309 Answer: \(3\)
Solution:
\(5^b\cdot5\) (essentially what you wrote except using multiplication) can be turned into \(5^{b+1}\). \(25^{b-1}\) can be turned into \(5^{2(b-1)}\), or \(5^{2b-2}\). Now you have to set these two equal to each other:
\(5^{b+1}\)= \(5^{2b-2}\)
You can get rid of the base, because they're the same.
\(b+1=2b-2\)
Subtracting \(b\) from both sides and adding 2 gives \(b=3\)
