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Aren't you supposed to find the inverse?

im just so confuse.... i have no idea what im doing

Is it written like this?
\[f(x) = (3-4x)^2, find f^{-1}(x)\]

yes!

ohh1! nope i honestly do not remember

wait you switch the x and y right?

ok so in my example would it be..
y = (3 – 4x)²
x = (3-4y)^2

so after the dust settles, what does y = ?

so x = (3-4y)^2
y= 1/4 (3-sqrt(x))
so... f^-1 = 1/4 (3-sqrt(x))

that's half of it...

Won't 1/4 (3 + sqrt(x)) = y also?

yes so it would be y = (1/4)( 3 ± sqrt(x))

Right. So, is the inverse a function?

f-1(x) = y = (1/4)(3 ± sqrt(x)) ??

oh no sorry its not a function!

oh okay! i see what your saying now! you explained it to me well! thank you for your help!! (:

Glad I could help make the little bulb come to life :-)

little light bulb that is

hahah you are awesome! can you maybe help me with one more?

I can try!

Use the Change of Base Formula to evaluate log7(76)

like would it be log7(76)= log(76)/log10(7) is that the answer?

\[\log_b a = \frac{\log a }{\log b}\]
So \[\log_7 76 = \frac{\log{76}}{\log 7}\]

wait so would it be log76 = 1.88081359
and log 7= 84509804 and you would divide those right?

Yes, but I think you are missing a decimal in your value of log 7...

oh yeah i did! so it would .84509804

so the answer would be 2.22556685

so my answer was right?

Yes!

ah thank you again!!!!!

\[\sqrt{7}(\sqrt{x}-7\sqrt{7}\]

Is that supposed to be \[\sqrt{7}(\sqrt{x}-7\sqrt{7})\] and is the goal to simplify it?

oh yes sorry! Multiply and simplify if possible.

Well, if you distribute it out, what do you get?

\[\sqrt{7x}-49\sqrt{x}\]

where did the \sqrt{x} next to the 49 come from?

i have no idea.. how do you distribute that?

and that last bit reduces to -49

\[a(b-c) = ab - ac, a = \sqrt{7}, b=\sqrt{x}, c = 7\sqrt{7}\]

wait whats the answer.. im lost

Oh I see what you were doing now! sorry