For what value of a does (one-seventh) Superscript 3 a + 3 Baseline = 343 Superscript a minus 1?
2 answers:
Answer:
a = 0
Step-by-step explanation:
I find a graphing calculator useful for such questions. It shows the solution to be a = 0. For the graph, we have rewritten the equation from
(1/7)^(3a+3) = 343^(a-1)
to
(1/7)^(3x+3) -343^(x-1) = 0 . . . . . this graphing calculator likes x for the independent variable
__
If you recognize that 343 is the cube of 7, you might solve this by taking logarithms to the base 7.
(7^-1)^(3a+3) = (7^3)^(a-1)
Equating exponents of 7*, we get ...
-(3a+3) = 3(a -1)
-3a -3 = 3a -3 . . . . . eliminate parentheses
0 = 6a . . . . . . . . . . . add 3+3a
0 = a . . . . . . . . . . . . divide by 6
_____
* Equating exponents of 7 is the same as taking logarithms to the base 7. Here, we use the rules of exponents ...
1/a^b = a^-b
(a^b)^c = a^(bc)
You might be interested in
Step-by-step explanation:
- Masses of the bodies are x and y
- Distance at which they are kept is r
- When two bodies are of masses x and y are placed at a distance r then the gravitational force acting on them is given as:
- Where G is the gravitational constant.