Algebra: 7^a * 7^b = 7^c, (2^a)^b = 64, 3^b/3^c = 1/9
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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?
.arthur ohlsten
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Re: Algebra
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
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
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Re: Algebra
arthur ohlsten said:
arthur ohlsten
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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
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