I Was able to solve this by randomly substituting values though my method is not a proper way to solve. X cannot be greater than 0 as otherwise the left-hand side (L.H.S) < 1 (since the denominator is greater than 1), and if X>0, R.H.S>3 Then X < 0. X also must be greater than -3 (X > -3) as we can not get a negative answer in the L.H.S (in the R.H.S we get a negative value if X < -3) by exponentiating a positive number. So we have: X < 0 X > -3 Now by substituting, we can get quite easily X = -1. This is not a proper method though especially when X is not an integer.
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1/(2^x)=x+3 , 2^x=1/(x+3)
taking ln , xln2=0-ln(x+3)=-ln(x+3)
x=-1 and x+3=2 result x=-1
I Was able to solve this by randomly substituting values though my method is not a proper way to solve.
X cannot be greater than 0 as otherwise the left-hand side (L.H.S) < 1 (since the denominator is greater than 1), and if X>0, R.H.S>3
Then X < 0.
X also must be greater than -3 (X > -3) as we can not get a negative answer in the L.H.S (in the R.H.S we get a negative value if X < -3) by exponentiating a positive number.
So we have:
X < 0
X > -3
Now by substituting, we can get quite easily X = -1.
This is not a proper method though especially when X is not an integer.
Very nice! ❤
...(1/2)^x
-1
Very nice! ❤