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Mathematics 22 Online
OpenStudy (anonymous):

Which logarithmic graph can be used to approximate the value of y in the equation 2y = 5?

OpenStudy (fibonaccichick666):

did you type that correctly? There is not need for logs in that eq

OpenStudy (anonymous):

http://gyazo.com/94ece37b12e563e25a60b4aa5e649dc1

OpenStudy (anonymous):

yes i did heres the graphs

OpenStudy (anonymous):

http://gyazo.com/87e0bee511b6f506f266bb9500bb2157

OpenStudy (fibonaccichick666):

well, solve your eq for y to begin

OpenStudy (anonymous):

how do i do that?

OpenStudy (fibonaccichick666):

you don't know how to solve 2y=5 for y?

OpenStudy (anonymous):

its 2^y=5

OpenStudy (fibonaccichick666):

that makes more sense now

OpenStudy (fibonaccichick666):

alright, so, in order to get y out of the exponent, we need to undo it. The opposite of the exponent function is the log function. now, there are different bases for logs, you can have log base anything. Here we can use log base 2 to undo, 2^y

OpenStudy (fibonaccichick666):

just remember we need to do this to both sides

OpenStudy (anonymous):

so 2*^2 ?

OpenStudy (fibonaccichick666):

not, quite. We write it like this \[log_2(2^y)=log_2(5)\]

OpenStudy (fibonaccichick666):

so now, we know that the log base 2 undo's the 2^y, so we are just left with y. and we get \[y=log_2(5)\] so now, you just plug it into your calculator and see which point marked is closer.

OpenStudy (fibonaccichick666):

but again, I think there is a type-o because none of those are a horizontal line.

OpenStudy (anonymous):

thats excatly how i copied it from the problem

OpenStudy (anonymous):

could it be the last graph? @FibonacciChick666

OpenStudy (fibonaccichick666):

well, what is log base 2 of 5? (you can use wolfram alpha to figure that out if you wish)

OpenStudy (fibonaccichick666):

just a couple of decimal places, not all of em

OpenStudy (anonymous):

it was the last graph

OpenStudy (fibonaccichick666):

I know that is the only option(even though this is a very poorly phrased question), I just wanted to make sure you could figure out that value.

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