Find the value of this equation

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Discussion Overview

The discussion revolves around finding the value of the expression \( (4)^x - \frac{1}{2}y \). Participants explore the interpretation of the equation, clarify notation, and discuss potential simplifications and transformations of the expression.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether the expression is \( 4^x - \frac{1}{2}y \) or \( 4^x - \frac{1}{2y} \).
  • It is noted that the values for \( x \) and \( y \) are not provided.
  • One participant proposes that the expression can be rewritten as \( 4^{x - \frac{y}{2}} = \frac{4^x}{2^y} = 2^{2x - y} \).
  • Another participant expresses confusion regarding the transformation to \( 2^{2x} \) and seeks clarification on the steps involved.
  • There is a discussion about the equivalence of \( \frac{1}{2}y \) and \( \frac{y}{2} \), with participants confirming this relationship.
  • One participant expresses understanding of the steps leading to \( \frac{2^{2x}}{2^y} \) but seeks further clarification on the derivation of \( 2^{2x} \) from \( 4^x \).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the expression or the values of \( x \) and \( y \). Multiple interpretations and approaches are presented, indicating ongoing uncertainty.

Contextual Notes

Participants express confusion over notation and the lack of specific values for variables, which complicates the discussion. The transformations and simplifications proposed depend on the interpretation of the original expression.

Who May Find This Useful

Individuals interested in algebraic expressions, mathematical transformations, or those seeking clarification on notation and variable interpretation may find this discussion relevant.

ai93
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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
 
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mathsheadache said:
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

$\chi$ $\sigma$
 
mathsheadache said:
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 $$4^x-\frac{1}{2}y$$ or $$4^x-\frac{1}{2y}$$?

Second, are you given values for $x$ and $y$?
 
MarkFL said:
First, is it $$4^x-\frac{1}{2}y$$ or $$4^x-\frac{1}{2y}$$?

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
 
$$4^{x-\frac{y}{2}} = \frac{4^x}{4^{\frac{y}{2}}} = \frac{4^x}{2^y} = \frac{2^{2x}}{2^y} = 2^{2x-y}$$

... take your pick ;)
 
skeeter said:
$$4^{x-\frac{y}{2}} = \frac{4^x}{4^{\frac{y}{2}}} = \frac{4^x}{2^y} = \frac{2^{2x}}{2^y} = 2^{2x-y}$$

... 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)
 
mathsheadache said:
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)

$$\frac{1}{2}y = \frac{y}{2}$$
 
skeeter said:
$$\frac{1}{2}y = \frac{y}{2}$$

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
 
$$4^x = (2^2)^x = 2^{2x}$$
 
  • #10
skeeter said:
$$4^x = (2^2)^x = 2^{2x}$$

I understand now thank you!

Does anyone know this topic so I can revise furthermore?
 

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