What is 27^(2x) = 9^(x-3)
do you need x?
Yes. Sorry I forgot to write that.
x=-3/2
what's \(3^2=?\) and, say what's \(3^3=?\)
Can you explain how to get that?
\(\large { {\color{brown}{ 27}}^{2x} = {\color{brown}{ 9}}^{x-3}\qquad \begin{cases} 3^2\to {\color{brown}{ 9}}\\ 3^3\to {\color{brown}{ 27}} \end{cases}\qquad thus\implies (3^3)^{2x}=(3^2)^{x-3} }\) see how to get "x" now?
Um...not exactly sure....Am I suppose to combine the 2x and x-3?
\(\large { {\color{brown}{ 27}}^{2x} = {\color{brown}{ 9}}^{x-3}\qquad \begin{cases} 3^2\to {\color{brown}{ 9}}\\ 3^3\to {\color{brown}{ 27}} \end{cases}\qquad thus\implies (3^3)^{2x}=(3^2)^{x-3} \\ \quad \\ 3^{3\cdot (2x)}=3^{2\cdot (x-3)}\impliedby \begin{array}{llll} \textit{same base, thus}\\\ \textit{exponents must also equal each other} \end{array}\\ \quad \\ 3(2x)=2(x-3) }\)
\(\large a^{whatever} = a^{whatever}\) means that whatever = whatever
Ohh! So then: 6x=2x-6 4x=-6 x=-2/3
Is that right?
yeap
hmmm actually wait
Thanks!
\(\bf 6x=2x-6\implies 4x=-6\implies x=-\cfrac{\cancel{6}}{\cancel{4}}\implies x=-\cfrac{3}{2}\)
Oh whoops! Thanks for catching that!
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