SOLUTION: log4 4^0.8=x b^x=y 4^x=4^0.8 x=√4^0.8 x=1.74 but when changing 4^1.74=11.16 and 4^.8=3.03 I can't seem to be doing this right since the exponents do not match the value.

Question 479051: log4 4^0.8=x
b^x=y
4^x=4^0.8
x=√4^0.8
x=1.74
but when changing 4^1.74=11.16 and 4^.8=3.03 I can't seem to be doing this right since the exponents do not match the value. please help
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
log4 4^0.8=x
b^x=y
4^x=4^0.8
----------
If you mean
then x = 0.8
-------------
RELATED QUESTIONS
I'm trying to solve the inequality: -3 < x < 0 so that a < 1/(x+4) < b The... (answered by scott8148)

(x-4)^2+8=0 (answered by josgarithmetic)

x-4+3(x-8)<0 (answered by kekegeter )

x/(x-4) - 3/8 =... (answered by Fombitz)

SOLUTION: log4 4^0.8=x b^x=y 4^x=4^0.8 x=&#8730;4^0.8 x=1.74 but when changing 4^1.74=11.16 and 4^.8=3.03 I can't seem to be doing this right since the exponents do not match the value.