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OpenStudy (anonymous):

Which of the following is the solution to the equation 25^2c = square root of 5^(4c + 14) ?

OpenStudy (zzr0ck3r):

\[25^{2c}=\sqrt{5^{4c+14}}\]?

OpenStudy (anonymous):

the 4c+14 is not in the square root

OpenStudy (zzr0ck3r):

\[25^{2c}=\sqrt{5}(4c+14)\]?

OpenStudy (zzr0ck3r):

so confused

OpenStudy (zzr0ck3r):

\[25^{2c}=\sqrt{5}^{(4c+14)?}\]

OpenStudy (anonymous):

yes

OpenStudy (zzr0ck3r):

ok \[25^{2c}=5^{2c+7}\] make sense?

OpenStudy (anonymous):

yes so far

OpenStudy (zzr0ck3r):

\[(5^2)^{2c}=5^{2c+7}\]

OpenStudy (anonymous):

allright

OpenStudy (zzr0ck3r):

\[5^{4c}=5^{2c+7}\] take log base 5 of both sides \[\log_5(5^{4c})=\log_5(5^{2c+7})\\4c\log_5(5)=(2c+7)\log_5(5)\\4c*1=(2c+7)*1\\4c=2c+7\\2c=7\\c=\frac{7}{2}\]

OpenStudy (anonymous):

Thank you :)

OpenStudy (zzr0ck3r):

np

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