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Mathematics 12 Online
OpenStudy (anonymous):

Can someone check my answer? What is the solution of the equation 2^x+1 = 5? Round your answer to the nearest ten-thousandth. 0.7565 1.3219** 2.3219 2.3974

OpenStudy (anonymous):

xlog2=log4 x=log4/log2 I dont have a calc so im just showing the work and you check

OpenStudy (er.mohd.amir):

2^x=5-1=4=2*2 2^x=2^2 wen base same then power is same x=2

OpenStudy (anonymous):

\[2^{x-1}=5\]

OpenStudy (anonymous):

Oh then yeah its (x-1)log2 = log 5 so x-1=log5/log2 x- (log5/log2) + 1

OpenStudy (anonymous):

wait is it x+1 or x-1? youve written two different questions

OpenStudy (anonymous):

x+1 @ Maisa101

OpenStudy (anonymous):

@Maisa101

OpenStudy (anonymous):

yeah so x= (log5/log2) -1

OpenStudy (anonymous):

just plug into calc and solve

OpenStudy (anonymous):

okay thanks

OpenStudy (er.mohd.amir):

U write Q is wrong 2^(x+1)=5 is right way to written.

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