View Full Version : Advanced Question Using Fractions and Negative Exponents

MathStudent2011

06-14-2011, 11:35 AM

I am having troubles trying to calculate ((1/2)(1+2x))^(-1/2).

The answer is 1/(2*sqrt[2x+1])

Can some one please tell me where im going wrong?

I start with (1/2)*(2x+1) = (2x+1/2)^(-1/2),

then have (2/2x+1)^(1/2),

then have sqrt [2/2x+1]^1,

then have sqrt[2]/sqrt[2x+1] = 1/sqrt[2x+1].

Where am I going wrong?

Denis

06-14-2011, 01:45 PM

I am having troubles trying to calculate ((1/2)(1+2x))^(-1/2).

The answer is 1/(2*sqrt[2x+1])

If that's the answer, then expression should be: (1/2)(1 + 2x)^(-1/2)

JeffM

06-14-2011, 04:41 PM

You need help with order of operations.

For example (1/2)(2x + 1) = (2x + 1)/ 2, not (2x + 1/2).

I suspect that is also what is wrong with your reasoning.

((1/2)(2x + 1))^(-1/2) is not the same as (1/2)(2x + 1)^(-1/2).

See http://www.freemathhelp.com/order-of-operations.html

MathStudent2011

06-14-2011, 11:11 PM

Sorry I made some typos. I get that your dividing the whole term by 2 rather than just the single integer. Is there anything else wrong with my methodology?

MathStudent2011

06-14-2011, 11:19 PM

Thankyou, I'll try that.

mmm4444bot

06-15-2011, 12:53 AM

The answer is 1/(2*sqrt[2x+1])

Then this exercise seems to be nothing more than: "Rewrite the expression (1/2)(1+2x)^(-1/2) in radical form."

Can some one please tell me where im going wrong?

Maybe you have not yet learned that the following two expressions are different forms of the same value (for all Real n > 0):

n^(1/2) = sqrt(n)

We call the expression on the lefthand side "exponential form" and on the righthand side "radical form".

Given one form, we can switch to the other.

Of course, you also need to know the meaning of a negative exponent.

n^(-1/2) = 1/sqrt(n)

Therefore, in your exercise, you need only switch the expression (1 + 2x)^(-1/2) to radical form and multiply the result by 1/2.

If the exercise involves anything else, you have not clearly posted it. 8-)

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