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Re: indices (115 Views)
Posted by: joejoe (IP Logged)
Date: September 16, 2009 10:07AM
You want to solve
(3 to the power t) x (9 to the power t+3) = 27 squared
Which is
3^t x 9^(t+3) = 27^2
-Expand 9^(t+3)
3^t x 9^t x 9^3 = 27^2
-Divide both sides by 9^3
3^t x 9^t = 27^2/9^3 = 729/729 =1
3^t x 9^t = 1
So t=0 because 3^0=1 and 9^0=1
You can double check the solution by subsituting t=0 into 3^t x 9^(t+3) = 27^2
(3 to the power t) x (9 to the power t+3) = 27 squared
Which is
3^t x 9^(t+3) = 27^2
-Expand 9^(t+3)
3^t x 9^t x 9^3 = 27^2
-Divide both sides by 9^3
3^t x 9^t = 27^2/9^3 = 729/729 =1
3^t x 9^t = 1
So t=0 because 3^0=1 and 9^0=1
You can double check the solution by subsituting t=0 into 3^t x 9^(t+3) = 27^2
| Subject | Written By | Posted |
|---|---|---|
 indices (159 Views)
|
dixie | 09/12/2009 02:08PM |
| Re: indices (115 Views) |
joejoe | 09/16/2009 10:07AM |
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