Question

2^2n = 16^-1

Answer 1

@prettyboy20202

Answer 2

Hint:
16^-1 = 1/16 = 1/2^4 = 2^-4
Can you now solve for n?

Answer 3

yes......................nooooooo

Answer 4

re write 16 as 2^4 then the equation becomes 2^2n = 2^-4 the bases will cancel and you will be left with 2n=-4, n becomes -2.

Answer 5

Compare
2^2n=2^-4
and hence solve for the equality of the exponents.

Answer 6

????????? not following

Answer 7

if 2^2n=2^-4
then compare exponents to get
2n=-4
or
n=-2
as Isaiah had shown above.

Answer 8

so -4 = -4

Answer 9

Yep, that's a check!

Answer 10

so the answer is -4

Answer 11

The answer is to find the value of n.
When solving 2n=-4, we get n=-2.

Answer 12

ok what about the next one that's wrong i believe its the 4th Q

Answer 13

the next one is tricky

Answer 14

\[\frac{1}{4}\log_2(16)+\frac{1}{2}\log_2(49)=\log_2(x)\]
one by one:

\[\log_2(16)=4\] since \(2^4=16\) and so \(\frac{1}{4}\log_2(16)=\frac{1}{4}\times 4=1\)

Answer 15

\[\frac{1}{2}\log_2(49)=\log_2(\sqrt{49}) = \log_2(7)\] so now the equation is
\[1+\log_2(7)=\log_2(x)\]

Answer 16

subtract \(\log_2(7)\) from both sides to get
\[\log_2(x)-\log_2(7)=1\] and now rewrite the left as
\[\log_2(\frac{x}{7})=1\] then rewrite in equivalent exponential form as
\[\frac{x}{7}=2^2=2\] so all that is left is to solve \(\frac{x}{2}=7\) which should be easy