Algebra--> Solve for x in terms of a and b: a=4bsquaredx (a=4b[squared]x)?

Question

Algebra--> Solve for x in terms of a and b: a=4bsquaredx (a=4b[squared]x)?

I need your help!!!! :)

Answer 1

First of all, you can use the ^ symbol to denote exponents. It is on the number 6 key of your keyboard. For example 3 squared would be written as 3^2.

So your equation is...

a = 4b^2x

Well since everything is multiplying the x, except for the a on the other side, simply divide by everything but the x.

a/(4b^2) = (4b^2x)/4b^2

Since 4b^2 divided by itself is 1, and 1*x =x

a/(4b^2) = x

Hope that helped
10 pts best answer?

Answer 2

The equation reads "resolve for x in terms of a and b". What this suggests is that we would like x on one area of the equation, and a and b on the different, alongside with any coefficients. Now, rewrite the equation slightly in yet another way: a = 4b^2x. the way its initially written, devoid of parentheses, potential it rather is the way it would examine. in basic terms b is "squared", no longer 4b. ok, are you able to work out the place to circulate? you want "x" remoted, so use your arithmetic operations (+, -, *, /). undergo in recommendations what you do to a minimum of one area of the equation you ought to do to the different (look at dividing "/" the two components by utilizing 4). save going you're very almost there. The final steps are staring you interior the face (smile)!

Answer 3

To get "X" on one side of the equation by itself, just divide both sides by 4b squared, so X = a/4b squared.

Answer 4

not very clear but is this what u meant A = 4B^ x^2 ?
^ = to the Power of.

A = 4B^ x^2

B^ X^2 = A/4

Ln( B^ X^2) = Ln(A/4)

X^2 ln(B) = ln(a/4)

X^2 = [ln(a/4)] / [ln(B)]

X^2 = ln [ (A/4) / B ]

X^2 = ln [ A/4B ]

X = Square root { ln [ A/4B] }


--------OR-------------------

A = 4 B^2 X

A / [ 4 B^2] = [4 B^2 X] / [ 4 B^2 ]

A / [ 4 B^2] = X

X = A / [ 4 B^2]