Question

5^log base 5 of 17

Answer 1

\[\huge\rm 5 \log_5 17\]
power rule \[\huge\rm log_b x^y = y \log_b x\]

Answer 2

Saying we have \(\Large A = 5^{\log_5(17)}\)

Take log of base 5 of both sides and you should have \(\Large \log_5(A) = \log_5(17)\log_5(5)\)

Since \(\log_5(5)=1\), you have \(\Large \log_5(A) = \log_5(17)\)

Now cancel \(\log_5\) out and you are left with \(\boxed{A=17}\)

Answer 3

hmm cancel the logs?
\[\log_b(x)\] and \(b^x\) are inverses
therefore it is always true that
\[\huge b^{\log_b(x)}=x\] and
\[\huge \log(b^x)=x\]

Answer 4

you are right for this equation