http://prntscr.com/6il22s please help!
am i correct??
x= -1 is correct.
thank you
perl
was he right??
yeah
yeah its ^-4
but i dont understand the ) after the 4
$$ \Large g(x) = 4^{-x} = \frac{1}{4^x} $$
no we have to solve for f(x) by substituing the ^-x with one of the other numbers on the side
$$ \Large{ g(x) = 4^{-x} = \frac{1}{4^x} \\ g(-1 ) =4^{-(-1)} = 4^{1} = 4 } $$
you can't really solve g(x) = f(x) algebraically . you can graph the two functions and find where they intersect.
hmm why won't logs work...
perl i think ur a bit too advanced xD
@domebotnos you can graph the two curves and find where they intersect https://www.desmos.com/calculator/cif42p0ppp
perl perl perl calm down this is wayyy to advanced
this is like solving for the x so i have to replace the x for the t=number provided
$$ \Large{ f(x) = g(x) \\ 3x+7 = 4^{-x}\\ \ln(3x+7) = \ln (4^{-x})\\ \ln(3x+7) = -x\cdot\ln (4)\\ \frac{\ln(3x+7)}{\ln 4} = -x } $$ this solution goes in circles.
PERL XD WHAT IS THAT
coconut asked why you cant used logs to solve f(x) = g(x)
what is logs???
there is a way to solve this using a special function called the Lambert function
give me one second let me show u
a log is the inverse of an exponential function. but you don't need to use logs here. if you fill out the table values, find the x value where the y values are equal
yeah that what i mean
this is the longest graph perl :p thank you !
$$\Large{ \begin{array}{|c|c|} \hline x & \text{f(x)=3x+7} & \text{g(x)}=4^{-x} \\ \hline -2 & 1& 16 \\ \hline -1 & 4 & 4 \\ \hline 0 & 7 & 1 \\ \hline 1&10 & \frac{1}{4}\\ \hline 2&13&\frac{1}{16}\\ \hline \end{array} } $$
ok then i was right! ok thank you soooooooooooooooooooooooo much perl!!!!
$$ \LARGE \color{blue}{Your~ welcome} $$
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