SOLUTION: given the following three equations, solve for a, b, and c:
7^a * 7^b = 7^c
(2^a)^b = 64
(3^b)/(3^c) = 1/9
thankz:D
Algebra.Com
Question 194947: given the following three equations, solve for a, b, and c:
7^a * 7^b = 7^c
(2^a)^b = 64
(3^b)/(3^c) = 1/9
thankz:D
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
given the following three equations, solve for a, b, and c:
7^a * 7^b = 7^c
(2^a)^b = 64
(3^b)/(3^c) = 1/9
:
Take each equation and simplify it:
:
Add exponents when you multiply
therefore we can say:
a + b = c
:
Multiply exponents when you raise it to another power
64 is the 6th power of two
therefore we can say:
ab = 6
:
we subtract the dividing exponent, 9 is the 2nd power of three
which is
therefore we can say:
b - c = -2
:
Rearrange and solve using elimination method on the 1st and 3rd equations
a + b - c = 0
0 + b - c = -2
----------------subtraction eliminates b and c, find a
a = +2
:
Find b using: ab = 6
2b = 6
b = 6/2
b = 3
:
Find c using b - c = -2
3 - c = -2
-c = -2 - 3
-c = -5
therefore
c = +5
;
;
Try these three solutions in each original equation.
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