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 solve exponential equation algebracally
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kingvg
New Member

USA
2 Posts

Posted - 03/22/2011 :  20:44:25  Show Profile  Reply with Quote


(1+ 0.10/12)12t=2
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Mrspi
Advanced Member

USA
998 Posts

Posted - 03/23/2011 :  20:48:43  Show Profile  Reply with Quote
quote:
Originally posted by kingvg



(1+ 0.10/12)12t=2



Since your subject line says "solve EXPONENTIAL equation algebraically," I wonder if you mean that 12t to be an exponent....


[1 + (0.10/12)]^(12t) = 2

I'd do the arithmetic inside the square brackets first: 1 + (0.10/12) = 1.00833 rounded to 5 decimal places.

(1.00833)^(12t) = 2

Now, take the log of both sides. You can use either common logs or natural logs. I'll use natural logs (ln):

ln (1.00833)^(12t) = ln 2

Use the rule of logs which says log a^m = m log a:

12t * ln (1.00833) = ln 2

t * 12 ln 1.00833 = ln 2

Divide both sides of the equation by (12 ln 1.00833):

t = (ln 2) / (12 ln 1.00833)

Now, it's calculator time....


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kingvg
New Member

USA
2 Posts

Posted - 03/24/2011 :  10:56:26  Show Profile  Reply with Quote
Yes, 12t is the exponent

Thanks
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