I will read your equation as
y = 2log2 (2(x+2)) - 4
so when x = 2 , we have
y = 2log2 (8) - 4 , so far the same as yours
y = 2(3) - 4
= 2
We seem to agree based on the way you typed the question.
Could you please verify and calculate the following logarithmic equation:
y = 2log_2(2)(x+2)-4 , in which x=2.
So the equation would become:
y = 2log_2(2)(2+2)-4
Solving...
y = 2log_2(2)(4)-4
y = 2log_2(8)-4
y = 2(3)-4
y = 6-4
y = 2
So I got y=2, however my answer sheet and my graphing calculator when I typed in the equation gave me y=4. How is this and where did I miscalculate?
Please note: that the "_" means subscript.
4 answers
Just tried different combinations of your typing ...
if you meant
y = 2log2 2^(x+2) - 4
then
y = 2log2 2^4 - 4
= 2log2 16 - 4
= 2(4) - 4
= 4
if you meant
y = 2log2 2^(x+2) - 4
then
y = 2log2 2^4 - 4
= 2log2 16 - 4
= 2(4) - 4
= 4
Thank you. I actaully just figured out that I indeed had miscalculated the equation. It was: y = 2log_2(2) (x+2)-4
Therefore I should solve the "log" part first:
1) y = 2log_2(2) (2+2)-4
2) y = 2 (2+2)-4
3) y = 2 (4)-4
4) y = 8-4
5) y = 4
In other words, the number "2" just before "x" applied to the 2log_2 and was not supposed to be multiplied with the (2+2) until the second step.
Therefore I should solve the "log" part first:
1) y = 2log_2(2) (2+2)-4
2) y = 2 (2+2)-4
3) y = 2 (4)-4
4) y = 8-4
5) y = 4
In other words, the number "2" just before "x" applied to the 2log_2 and was not supposed to be multiplied with the (2+2) until the second step.
No, wait, actually the 2 does not apply to the 2log_2 by itself. Rather, the "2(x+2)" applies to the log. I was so confused, sorry. So I'm going to assume that the answer "2" is correct and the "4" on my answer sheet is incorrect.