To solve for x I thought I would rewrite the equation in exponential form:
3^(2-x) = 3
Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3
Why is this incorrect?
To solve for x I thought I would rewrite the equation in exponential form:
3^(2-x) = 3
Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3
Why is this incorrect?
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log_3(2 - x) = 3
3^log_3(2 - x) = 3^3
2 - x = 9
x = -7
Last edited by apricissimus; 08-03-08 at 06:19 PM.
I actually solved that a short bit ago. Thanks though.
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Uh...3³ is not 9. It's 27.
Okay okay, now how about this:
The best I can do is rewrite it as .5xLog(.7) = 5log(1-x)
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Is this your math hw?
Yes, and let me write my disclaimer now:
I seek guidance only, not the answer.
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Just plug it into your calculator's numeric solver and be done with it.
That's not correct. You did the left side of the equation correctly, but you got things backwards on the right side.
Remember, log(x^y) = y* log(x)
Oh, so I should have
.5xlog(.7) = 1-xlog(5)
then divide by log5 on both sides to begin isolation of the variable?
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By some algebraic manipulation, you can rewrite the equation as (35/2)^x = 5^2. (Try to do it yourself.)
From here take the log (to a certain base b) of both sides, and use the change of base formula, log_b(x) = log_a(x)/log_a(b) (or ln(x)/ln(b) if you like natural logarithms. And who doesn't?).