MHB How Does Simplifying Exponential Expressions Work?

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The discussion focuses on simplifying the exponential expression \(\left(\frac{1}{2}\right)^{2-\frac{1}{2}\log_2(9)}\). Participants clarify the expression and demonstrate the simplification process, ultimately showing that it simplifies to \(\frac{3}{4}\). Various methods are presented, including using logarithmic properties to transform the expression. The conversation emphasizes the importance of understanding logarithmic identities in simplifying exponential forms. The final result confirms that the simplified expression equals \(\frac{3}{4}\).
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View attachment 3134 sorry for posting like this my computer broke down. I have trouble with this task
 

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Alexstrasuz said:
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 $$\left ( \frac{1}{2} \right ) \cdot 2-12 \log_2 9$$ ??
 
View attachment 3135
This is what I ment
EDIT:2-1/2log29
Is exponent of 1/2
 

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

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

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