Simple Algebra Question: Finding the Value of X in Equations

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Homework Help Overview

The discussion revolves around algebraic equations encountered in a physics context, specifically focusing on finding the value of x in two equations. The first equation involves manipulating powers and roots, while the second equation involves logarithmic properties.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss isolating terms and manipulating equations to solve for x. There are attempts to use logarithms and algebraic manipulation, with some questioning the correctness of their approaches. One participant expresses confusion about canceling terms in logarithmic equations.

Discussion Status

Some participants have shared their attempts and partial solutions, while others are still exploring different methods to approach the second equation. There is acknowledgment of a correct solution for the first problem, but uncertainty remains regarding the second equation.

Contextual Notes

Participants note that the discussion is occurring in a physics forum, which may influence the framing of the algebraic problems. There is mention of a need to combine terms algebraically and the challenge of handling logarithmic expressions.

bengaltiger14
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I know this is not an algebra forum but this problem arose for me in physics.

Determine a numerical value for x:

((9^3)/27) = ((SQRT(24^4))/(3^x))

I move the 27 to the top and the 3^x to the top and then solve for x this way.

3 - 1 = 2 - x, this makes x equal to 0 but it is not right. What is the correct way to do this?

Also, determine x for: 3^2x = 7(3^x)

I used logs: 2xlog3=7(xlog3) The xlog3 will cancel and I can't solve for x anymore. Any recommendations?
 
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bengaltiger14 said:
I know this is not an algebra forum but this problem arose for me in physics.

Determine a numerical value for x:

((9^3)/27) = ((SQRT(24^4))/(3^x))

I move the 27 to the top and the 3^x to the top and then solve for x this way.

I would isolate the 3x to one side of the equation by itself. Do you see what to do then?

3 - 1 = 2 - x, this makes x equal to 0 but it is not right. What is the correct way to do this?

Also, determine x for: 3^2x = 7(3^x)

I used logs: 2xlog3=7(xlog3) The xlog3 will cancel and I can't solve for x anymore. Any recommendations?

Try combining the terms algebraically first so that you only have one x in the equation.
 
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I solved the first problem and got 3, which is the correct answer.. Thanks.

I am still having problems with the next one. Do you divide both sides by 3^x and solve for x
 
bengaltiger14 said:
I solved the first problem and got 3, which is the correct answer.. Thanks.

I am still having problems with the next one. Do you divide both sides by 3^x and solve for x

Yes, that's what I would do.
 

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