Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png :http://webwork.mathstat.concordia.ca...d000b7e941.png

I made it into

log_{2}x^{9/2}=-3

2^{-3}=x^{9/2 .125x-4.5=x -0.5625=x but I don't think it's right}

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- October 8th 2012, 06:44 AMM670Logs
Solve for http://webwork.mathstat.concordia.ca...dd0b8b8e91.png :http://webwork.mathstat.concordia.ca...d000b7e941.png

I made it into

log_{2}x^{9/2}=-3

2^{-3}=x^{9/2 .125x-4.5=x -0.5625=x but I don't think it's right} - October 8th 2012, 06:52 AMskeeterRe: Logs

- October 8th 2012, 07:09 AMHallsofIvyRe: Logs
Skeeter's method, dividing both sides by 9/2 first, is simpler than yours but it can be done your way:

I don't know how you now got ".125x-4.5=x". Was that first "x" supposed to be a multiplication sign? Bad idea! Never use the same symbol to mean two different things!

But even so, you get rid of the "9/2" power on the left by taking the 9/2 root- which is the same thing as the 2/9 power, on both sides:

If you want or are required to "rationalize the denominator", multiply both numerator and denominator by : . - October 8th 2012, 07:12 AMSorobanRe: Logs
Hello, M670!

Quote:

Solve for

I made it into:

. . Right!

. . Correct!

. . I don't understand this step.

. . Nor this step.

but I don't think it's right

In your second step, you had:

Raise both sides to the power:

. . .

Therefore: .

- October 8th 2012, 04:19 PMM670Re: Logs
sorry guys I should of written my step .125x-4.5=x as follows .125(-4.5)=x

but i understand what you have all shown me