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Date Posted:
3 years ago
Thank you sir
Thank you sir
One more question sir
Solve x= (4x)^lg x
Solve x= (4x)^lg x
x = (4x)^lg x
Taking lg on both sides,
lg x = lg (4x)^lg x
We can bring down the power “lg x” using the power rule. Note that the power rule works because of the lg base we just introduced, not the original lg x in the power.
lg x = (lg x) (lg 4x)
(lg x) (lg 4x) - lg x = 0
(lg x) (lg 4x - 1) = 0
lg x = 0 or lg 4x = 1
x = 10^0 or 4x = 10^1
x = 1 or x = 5/2
Taking lg on both sides,
lg x = lg (4x)^lg x
We can bring down the power “lg x” using the power rule. Note that the power rule works because of the lg base we just introduced, not the original lg x in the power.
lg x = (lg x) (lg 4x)
(lg x) (lg 4x) - lg x = 0
(lg x) (lg 4x - 1) = 0
lg x = 0 or lg 4x = 1
x = 10^0 or 4x = 10^1
x = 1 or x = 5/2
Thank you sir
Can i divide both sides by lgx
Nope, because cancelling lg x on both sides will also cancel out a very critical solution lg x = 0.