Algebra: 7^a * 7^b = 7^c, (2^a)^b = 64, 3^b/3^c = 1/9

[Tai]

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

[mmm4444bot]

Use the Properties of Exponents to write equations where a, b, and c are no longer exponents.

I'll get you started.

3^b/3^c = 3^(b-c)

We're told that these expressions equal 1/9.

We know that 3^(-2) = 1/3^2 = 1/9

Therefore:

3^(b - c) = 3^(-2)

By equating the exponents above, we get the following relationship.

b - c = -2

Does this make sense?

[Tai]

it makes sense... THANKS ALOT :).

[arthur ohlsten]

7^a*7^b=7^c
7^(a+b)=7^c
eq 1 a+b=c

[2^a]^b=64
2^(ab)=64
eq 2 ab=6

3^b/3^c=1/9
3^(b-c)=1/9
eq3 b-c=-2

from eq3 c=b-2 substitute into eq1
a+b=b-2
a=-2 answer

from eq2 ab=6 but a=-2
b=-3 answer

from eq 1 a+b=c
-2-3=c
c=-5

arthur

[Subhotosh Khan]

7^a*7^b=7^c
7^(a+b)=7^c
eq 1 a+b=c

[2^a]^b=64
2^(ab)=64
eq 2 ab=6

3^b/3^c=1/9
3^(b-c)=1/9
eq3 b-c=-2

from eq3 c=b-2 substitute into eq1 <<< Incorrect ... it should be c = b + 2
a+b=b-2
a=-2 answer

from eq2 ab=6 but a=-2
b=-3 answer

from eq 1 a+b=c
-2-3=c
c=-5

a = -2, b = -3 and c = -5 do not satisfy the original equations
arthur

[arthur ohlsten]

1) a+b=c
2)ab=6
3)c-b=2

from 1and 3
a=c-b
2=c-b subtract
a-2=0
a=2

from 2)
b=3

from 1)
c=5

a=2 b=3 c=5
I goofed with the signs when I wrote the answers





1)
7^a*7^b=7^c
7^2 * 7^3 =7^5
[49][343]= 16,807
16807=16807

a=2 b=3 c=5 satisfys first equation

2)
[2^a]^b=64
[2^2]^3=64
[4]^3=64
64=64
a=2b=3c=5 satisfys the second equation

3)
3^b/3^c=1/9
3^3 /3^5 = 1/9
27/243 =1/9
1/9=1/9
a=2 b=3 c=5 satisfys the 3rd eq.
Arthur