Find the value of this equation

[ai93]

My question is find the value of (4)^x-1/2y
Sorry I just joined and not sure how to use the symbols. Also I would try and show my workings out but i am stumped! Need some help

 

[chisigma]

My question is find the value of (4)^x-1/2y
Sorry I just joined and not sure how to use the symbols. Also I would try and show my workings out but i am stumped! Need some help

Welcome on MHB mathsheadache!...

... usually an equation in one unknown x is written in form of equality like f(x)=0 and his solution is to find the values ​​of x that satisfy the equality...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

My question is find the value of (4)^x-1/2y
Sorry I just joined and not sure how to use the symbols. Also I would try and show my workings out but i am stumped! Need some help

First, is it \(\displaystyle 4^x-\frac{1}{2}y\) or \(\displaystyle 4^x-\frac{1}{2y}\)?

Second, are you given values for $x$ and $y$?

 

[ai93]

First, is it \(\displaystyle 4^x-\frac{1}{2}y\) or \(\displaystyle 4^x-\frac{1}{2y}\)?

Second, are you given values for $x$ and $y$?

x and y has no values

It would be the first one, the 4 is in brackets and it would all be to the power of 4.
I hope you understand

 

[skeeter]

\(\displaystyle 4^{x-\frac{y}{2}} = \frac{4^x}{4^{\frac{y}{2}}} = \frac{4^x}{2^y} = \frac{2^{2x}}{2^y} = 2^{2x-y}\)

... take your pick ;)

 

[ai93]

\(\displaystyle 4^{x-\frac{y}{2}} = \frac{4^x}{4^{\frac{y}{2}}} = \frac{4^x}{2^y} = \frac{2^{2x}}{2^y} = 2^{2x-y}\)

... take your pick ;)

I am not sure how you got -y/2. The question would be X minus a half as a fraction then Y all to the power of (4)

 

[skeeter]

I am not sure how you got -y/2. The question would be X minus a half as a fraction then Y all to the power of (4)

\(\displaystyle \frac{1}{2}y = \frac{y}{2}\)

 

[ai93]

\(\displaystyle \frac{1}{2}y = \frac{y}{2}\)

Oh wow that makes a lot more sense! So can you say Y x 1/2 would equal Y/2?
and I understand the steps until 2^2x/2Y. Where did the 2^2x come from?

Thank you

 

[skeeter]

\(\displaystyle 4^x = (2^2)^x = 2^{2x}\)

 

[ai93]

\(\displaystyle 4^x = (2^2)^x = 2^{2x}\)

I understand now thank you!

Does anyone know this topic so I can revise furthermore?