Question

Which of the following is the solution to the equation 25^(2c) = square root of 5^(4c + 16) ? Select one:
A. c = -5 over 2
B. c = 7 over 2
C. c = 4
D. c = -7 over 2

Answer 1

check your equation, are the terms as \(
25^{2c}= \sqrt{5^{(4c+16)}}
\) ? cuz I get a funky answer from that

Answer 2

That's what the test says, this one got me too confused.

Answer 3

ok, lemme check some more

Answer 4

okay

Answer 5

I just found my error, ok, give a few secs

Answer 6

Okay

Answer 7

the one thing you have to bear in mind is that, whenever you want to get an EXPONENT DOWN, you use a Logarithm Cancellation rule, where the Log base is the same as the coefficient, lemme type it

Answer 8

okay

Answer 9

$$
25^{2c}= \sqrt{5^{(4c+16)}} \implies (5^2)^{2c}=\sqrt{5^{(4c+16)}}\\
\text{square both sides to get the expression OFF the root}\\
\pmatrix{(5^2)^{2c}}^2=\pmatrix{\sqrt{5^{(4c+16)}}}^2\\
5^{8c}=5^{(4c+16)}
$$

Answer 10

$$
log_5{5^{8c}}=log_5{5^{(4c+16)}}\\
\text{log cancellation rule, base 5, to get the exponents down}\\
8c=4c+16
$$
in this case both sides are of base 5, that is the NUMBER with the exponent is 5, thus, a log(5) gets both exponents

Answer 11

and from there, I'd guess you know what to do :)

Answer 12

Ans choice is C

Answer 13

so then \[4c=16\] and then simplified it's \[c=4\] right?

Answer 14

:D

Answer 15

Thanks for the help!

Answer 16

yw