MHB How Does Simplifying Exponential Expressions Work?

[Alexstrasuz1]

View attachment 3134 sorry for posting like this my computer broke down. I have trouble with this task

 

[mathmari]

View attachment 3134 sorry for posting like this my computer broke down. I have trouble with this task

Do you mean $$\left ( \frac{1}{2} \right )^2-12 \log_2 9$$ or $$\left ( \frac{1}{2} \right ) \cdot 2-12 \log_2 9$$ ??

 

[Alexstrasuz1]

View attachment 3135
This is what I ment
EDIT:2-1/2log₂9
Is exponent of 1/2

 

[anemone]

View attachment 3135
This is what I ment
EDIT:2-1/2log₂9
Is exponent of 1/2

Hi Alexstrasuz,

Okay, you have an expression that read $\left(\dfrac{1}{2}\right)^{2-\dfrac{1}{2\log_2 9}}$, what do you want to do about it? Do you mind to give us the exact working of the original problem

 

[mathmari]

View attachment 3135
This is what I ment
EDIT:2-1/2log₂9
Is exponent of 1/2

$$\left ( \frac{1}{2} \right )^{2-\frac{1}{2} \log_2 9} \ \ \ \overset{x \log y=\log y^x}{=} \\ \left ( \frac{1}{2} \right )^{2- \log_2 9^{\frac{1}{2}}}= \left ( \frac{1}{2} \right )^{2- \log_2 3}=\left ( \frac{1}{2} \right )^{2\log_2 2- \log_2 3}=\left ( \frac{1}{2} \right )^{\log_2 2^2- \log_2 3}=\left ( \frac{1}{2} \right )^{\log_2 4- \log_2 3} \ \ \ \overset{\log x - \log y=\log \frac{x}{y}}{=} \ \ \ \left ( \frac{1}{2} \right )^{\log_2 \frac{4}{3}}=\frac{1}{2^{\log_2 \frac{4}{3}}} \ \ \ \overset{x^{\log_x y}=y}{=} \ \ \ \frac{1}{\frac{4}{3}}=\frac{3}{4}$$

 

[MarkFL]

Another way to proceed:

$$\left(\frac{1}{2}\right)^{2-\frac{1}{2}\log_2(9)}=2^{\log_2\left(\frac{3}{4}\right)}=\frac{3}{4}$$

 

[soroban]

Hello, Alexstrasuz!

Simplify: .\left(\frac{1}{2}\right)^{2-\log_2(9)}

We have: .\left(\frac{1}{2}\right)^2\cdot\left(\frac{1}{2}\right)^{-\frac{1}{2}\log_2(9)} \;=\;\frac{1}{2^2}\cdot 2^{\frac{1}{2}\!\log_2(9)} . **

\;=\;\frac{1}{4}\cdot 2^{\log_2(9^{\frac{1}{2}})} \;=\;\frac{1}{4}\cdot 2^{\log_2(3)}\;=\;\frac{1}{4}\cdot 3 \;=\;\frac{3}{4}** . Note that: .\left(\frac{a}{b}\right)^{-n} \;=\;\left(\frac{b}{a}\right)^n