Question

Help me with logarithms! Please?

Answer 1

\[a^1=a \text{ which means } \log_{a}(a)=1 \\ a^0=1 \text{ which means } \log_{a}(1)=0 \\ \text{ in general } \\ a^x=y \text{ means } \log_a(y)=x\]

Answer 2

Ok. I got that much.

Answer 3

great do you have anymore questions?

Answer 4

this what the function roughly looks like

Answer 5

Yea, but you see how do i know which point it passes through. Look at the answer choices to the second drop down on the attachment

Answer 6

oh I thought you understood that part because you said you got that much

Answer 7

Well, i understood how you got it for the first part, how about the second drop down.

Answer 8

\[f(x)=\log_a(x) \\ y=f(x) \\ \text{ do you see above we have } \\ \log_a(a)=1\]
that last equation says when x is a , y is 1

Answer 9

f(x) = loga(x), a>1 passes through (a^-1, -1), (1,0), (a, 1)
f(x) = loga(x), 0<a<1 passes through (a, 1), (1, 0), (a^-1, -1)

Answer 10

Ohhhhhh. okay. It makes some sense now.

Answer 11

\[f(x)=\log_a(x) \\ \text{ in general say you plug in } m \text{ and you get } n \\ \log_a(m)=n \text{ we say the ordered pair } (m,n) \text{ is on } y=\log_a(x)\]

Answer 12

Alrightie. Thanks.

Answer 13

for the first rule,

\(\large\color{black}{ \displaystyle a^1=a~~\Rightarrow~ \log_a(a)=1 }\)
this is true when \(\large\color{black}{ \displaystyle a>0,~~a\ne\left\{0,1\right\} }\)


for the second rule
\(\large\color{black}{ \displaystyle a^0=1~~~\Rightarrow~~~ \log_a(1)=0~,~~~~a>0 }\)

but everything else is good. If I made a little error by accident, please correct me.