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Mathematics 19 Online
OpenStudy (anonymous):

if log5(x+2)=2, then x=?

OpenStudy (anonymous):

23

OpenStudy (anonymous):

how did you get that?

OpenStudy (anonymous):

let x+2 = y log5 y = 2 hence, y = 5^2 = 25 x+2=25 x=23

OpenStudy (anonymous):

\[\log_{5}(x+2)=2 \] Can be rearranged in exponential form:\[(x+2) = 5^2\] \[x = 5^2 - 2\] \[x = 23\]

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