Algebra

[IloveManUtd]

In my question:

Solve 2 [ 9^ (x-1)] - 5 (3^x)= 27

If I change 9 to be 3^2, do I put the square as 3(^2x) or what? Thanks.

 

[ryan_kalle]

No expert here but I believe 9^(x-1) will be 3^[2(x-1)].

 

[IloveManUtd]

No expert here but I believe 9^(x-1) will be 3^[2(x-1)].

So how do I subsitute 3^[2(x-1)] with an unknown, say "u" ?

 

[soroban]

Hello, IloveManUtd!

This requires some Olympic-level gymnastics . . .

Solve: . \(\displaystyle 2\!\cdot\!9^{x-1} - 5\!\cdot\!3^x \:=\: 27\)

\(\displaystyle \text{The first term is: }\;2\!\cdot\!9^{x-1} \;=\;2\!\cdot\!(3^2)^{x-1} \;=\;2\!\cdot\!3^{2(x-1)} \;=\;2\!\cdot\!3^{2x-2}\)

. . . . . . . . . . . \(\displaystyle =\;2\!\cdot\!3^{2x}\!\cdot\!3^{-2} \;=\;2\!\cdot\!3^{2x}\!\cdot\!\frac{1}{9} \;=\;\frac{2}{9}\!\cdot\!3^{2x} \;=\;\frac{2}{9}\!\cdot\!\left(3^x\right)^2\)


\(\displaystyle \text{The equation becomes: }\:\frac{2}{9}\!\cdot\!\left(3^x\right)^2 - 5\!\cdot\!3^x - 27 \;=\;0\)

\(\displaystyle \text{Multiply by 9: }\;2\!\cdot\!\left(3^x\right)^2 - 45\!\cdot\!3^x - 243 \;=\;0\)


\(\displaystyle \text{Let: }\,u \:=\:3^x\)

\(\displaystyle \text{Substitute: }\;2u^2 - 45u - 243 \;=\;0\)

\(\displaystyle \text{Factor: }\;(2u+9)(u=27) \;=\;0\)

\(\displaystyle \text{Hence: }\:u \:=\:-\frac{9}{2},\;27\)


\(\displaystyle \text{Back-substitute:}\)

. . \(\displaystyle 3^x\:=\:-\frac{9}{2} \quad\hdots\;\text{ no real roots}\)

. . \(\displaystyle 3^x \:=\:27 \quad\Rightarrow\quad \boxed{x \:=\:3}\)

 

[IloveManUtd]

Thanks man, soroban. U're a great help!

 

[lookagain]

\(\displaystyle \text{Factor: }\;(2u+9)(u=27) \;=\;0\)

\(\displaystyle \rightarrow \text{Hence: }u = (2u + 9)(u - 27)=0\) . . . . . Correction (for the typo)

\(\displaystyle \text{Hence: }\:u = -\frac{9}{2},\;27\)