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 
:
Take each equation and simplify it:
:
{{{7^a * 7^b = 7^c}}}
Add exponents when you multiply
{{{7^(a+b) = 7^c}}}
therefore we can say:
a + b = c
:
{{{(2^a)^b = 64}}}
Multiply exponents when you raise it to another power
{{{2^(ab) = 64}}}
64 is the 6th power of two
{{{2^(ab) = 2^6}}}
therefore we can say:
ab = 6
:
{{{(3^b)/(3^c) = 1/9 }}}
we subtract the dividing exponent, 9 is the 2nd power of three
{{{3^(b-c)= 1/3^2 }}}
which is
{{{3^(b-c) = 3^-2 }}}
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.