Graph 2^y=3
@nelsonjedi @johnweldon1993 @HelpBlahBlahBlah @hero
You need to determine the value for y. You will have to convert to a logarithmic equation I assume the question is\[2^{y}=3\]
@nelsonjedi Yes
So the logarithmic equation would look like what?
log(6)
@dan815
log(6x)
Does this look familar?\[\log_{2} 3 =y\]
Yeah but how do I graph it @nelsonjedi
Solve for y which will be (log 3) / (log 2). It will be the y value which is constant and thus will be a straight line.
None of the choices are a straight line
Which logarithmic graph can be used to approximate the value of y in the equation 2^y = 3? @nelsonjedi
What are the choices?
How is that possible with no X variable in the equation?
That is why I am confused, the problem seems to be incorrect, and it is worth like half of the test, what do I do, I am on online school
@gamer456148 maybe if you could post a screenshot of the material, so we can see it ?
other than that nelsonjedi is correct, there's no "x", thus "y" is just a flat value
I could take a stab at it. Given that the (log3) / (log2) = 1.5849... Then maybe you should look at a graph that includes the point (3, 1.5849) Which looks like c?
What are the explicit equation and domain for an arithmetic sequence with a first term of 5 and a second term of 3? (5 points) an = 5 - 3(n - 1); all integers where n ≥ 1 an = 5 - 3(n - 1); all integers where n ≥ 0 an = 5 - 2(n - 1); all integers where n ≥ 0 an = 5 - 2(n - 1); all integers where n ≥ 1 Is this A or B, I am not sure What are the explicit equation and domain for a geometric sequence with a first term of 4 and a second term of -12? (5 points) an = 4(-3)n - 1; all integers where n ≥ 1 an = 4(-3)n - 1; all integers where n ≥ 0 an = 4(36)n - 1; all integers where n ≥ 1 an = 4(36)n - 1; all integers where n ≥ 0
@nelsonjedi @jdoe0001
hmm that doesn't look like a screenshot
either way, tis better to post anew, thus more visibility since we may not know, and we can also revise each other
Did it
Join our real-time social learning platform and learn together with your friends!