Answer:
2. w = -3
4. c = -2
6. r = 7
8. n = 4
Step-by-step explanation:
It's hard to read the photo, but I'll try.
2.
[tex] 2^{w + 4} \times 2^{4w + 6} = 2^{2w + 1} [/tex]
On the left side, you have a product of 2 powers with the same base.
Give the same base and add the exponents, as in the rule
[tex] a^m \times a^n = a^{m + n} [/tex]
You get on the left side 2 to the sum of the exponents.
[tex] 2^{w + 4 + 4w + 6} = 2^{2w + 1} [/tex]
Simplify the long exponent on the left side.
[tex] 2^{5w + 10} = 2^{2w + 1} [/tex]
Now we have another rule we can use. If two powers are equal, and their bases are equal, then the exponents must be equal.
Here, 2 to the power 5w + 10 equals 2 to the power 2w + 1. Since both bases are 2 and are equal, then the exponents must be equal.
5w + 10 = 2w + 1
Subtract 2w from both sides. Subtract 10 from both sides.
3w = -9
Divide both sides by 3.
w = -3
4.
[tex] \dfrac{1}{5} = 5^{2c + 3} [/tex]
We need to write the left side as a power of 5. Then we equate the exponents like we did in problem 2.
Recall the rule:
[tex] a^{-n} = \dfrac{1}{n} [/tex]
Apply this rule to the left side in reverse.
[tex] 5^{-1} = 5^{2c + 3} [/tex]
Now we have two powers that are equal, both having the same base, 5, so the exponents must be equal.
2c + 3 = -1
2c = -4
c = -2
6.
[tex] 216 = 6^{2r - 11} [/tex]
This is the same idea as problem 4. Write the left side as a power of 6.
[tex] 6^3 = 6^{2r - 11} [/tex]
[tex] 2r - 11 = 3 [/tex]
[tex] 2r = 14 [/tex]
[tex] r = 7 [/tex]
r = 7
8.
[tex] 4^{n} \times 4^{2n - 9} = 64 [/tex]
This problem is similar to problem 2.
[tex] 4^{n + 2n - 9} = 4^3 [/tex]
[tex] 4^{3n - 9} = 4^3 [/tex]
3n - 9 = 3
3n = 12
n = 4
