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Mathematics 6 Online
OpenStudy (anonymous):

find y in terms of x when log2y=2+log2x

OpenStudy (lgbasallote):

do you know how to change log_2 y = 2 + log_2 x into exponential form?

OpenStudy (anonymous):

need to find y in terms of x when log2y=2+log2x

OpenStudy (anonymous):

did you ask me a question? It's my first time here, so I'm not familiar with right steps to follow

OpenStudy (lgbasallote):

do you know how to change log_2 y = 2 + log_2 x into exponential form?

OpenStudy (lgbasallote):

i'm asking if you know how to start

OpenStudy (lgbasallote):

..how to start solving the problem

OpenStudy (anonymous):

wouldn't yo uconvert this to exponential form, so y=2^(2+log2x)?

OpenStudy (lgbasallote):

right

OpenStudy (lgbasallote):

now.. are you familiar with this concept: \[\huge a^b \times a^c = a^{b+ c}\]

OpenStudy (anonymous):

yes

OpenStudy (lgbasallote):

good. so can you apply that in 2^(2+ log_2 x)

OpenStudy (anonymous):

could you convert the log_2 x into exponential form also?

OpenStudy (lgbasallote):

tell me what you got for 2^(2+log_2 x) first

OpenStudy (anonymous):

i'm not entirely sure on that one

OpenStudy (lgbasallote):

use the property a^(b+c) = a^b * a^c

OpenStudy (anonymous):

okay well than it should be (2^2)*(2^log_2 x)

OpenStudy (lgbasallote):

right

OpenStudy (lgbasallote):

now here comes another property \[\huge a^{\log_a b} = b\] do you understand what that means?

OpenStudy (anonymous):

yes - does that mean that the answer to this question would be 4x?

OpenStudy (lgbasallote):

right. but the complete answer is y = 4x

OpenStudy (anonymous):

because you want the expression in terms of y

OpenStudy (anonymous):

alright, that makes sense - thanks a lot for the help!

OpenStudy (anonymous):

Is this site always free or do I have to sign up for a fee if I need more help?

OpenStudy (lgbasallote):

no. it's very free

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