Question

Evaluate log (base 9) 27

Answer 1

3/2

Answer 2

9^x=27
3^(2x)=3^3
2x=3 (a theorem allows this)
x=3/2

Answer 3

\[\log_{9}27 \]mean to what power do i have to raise 9 to make it equal 27

Answer 4

so we can write\[9^x=27\]

Answer 5

9 and 27 can both be written with base 3\[3^{2x}=3^3\]

Answer 6

there is a theorem:\[\log_{b}u=\log_{b} v \iff u=v\]and\[b^u=b^v \iff u=v\]

Answer 7

we use the second part to equate the exponents\[2x=3\]and then solve for x\[x=\frac{3}{2}\]done