Question

How do you get rid of the radical when it has a exponent attached like this?
Question below in a minute.

Answer 1

\[25^{2c} = \sqrt{5}^{4c+16}\]

Answer 2

i assume it is either:
\[25^{2c} = (\sqrt{5})^{4c+16}\]
or
\[25^{2c} = \sqrt{5^{(4c+16)}}\]

Answer 3

Can you show me how to solve it from here too?

Answer 4

only if we can determine which option you are refering to, but it should be doable either way.

Answer 5

one rule that might be useful to recall:
\[(n^a)^b=n^{a*b} \]

Answer 6

@CarrionBeast what option are you referring to? @amistre64 pointed out two possible problems, which one is it?

Answer 7

Option number one.

Answer 8

both forms
\[ 25^{2c} = (\sqrt{5})^{4c+16} \\
25^{2c} = \sqrt{5^{4c+16}} \]
can be written as
\[ 25^{2c} = 5^{\frac{4c+16}{2}} = 5^{2c+8} \]

Answer 9

I would write 25 as 5^2

Answer 10

alternatively, you could write
\[ 5^{2c+8}= 5^{2(c+4)} = \left( 5^2\right) ^{c+4} \\ = 25^{c+4} \]
and solve
\[ 25^{2c}= 25^{c+4} \]

by equating the exponents

Answer 11

btw, we use the idea that a square root is the same as using the exponent ½

Answer 12

Thank you