A few days ago
Maths problem: find x?
16^x – 12(4^x) + 32 = 0
Top 3 Answers
A few days ago
Favorite Answer
trust me the second answer is wrong……he got lucky becos the answer is 1. the right approach is write 16^x as 4^2x or (4^x)^2,
(4^x)^2-12(4^x)+32=0
then u substitute 4^x as lets say y
then we have y^2-12y+32=0
now solve it y^2-8y-4y+32=0
y(y-8)-4(y-8)=0
(y-8)(y-4)=0
Y=4 or 8
therefore asy=4^x=4 or 8
4^x=4, x=1
or
4^x=8
take log on both sides
log 4^x=log 8
xlog4=log8
x=log8/log4=1.5
therefore x=1 and 1.5
u can check it if you want to
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A few days ago
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A few days ago
First you need to do some simplifying:
16^x – 12(4^x) + 32 =0 would be….
16^x – 48^x + 32 =0 which simplifies to….
-32^x + 32 =0 which leads us to ….
x = 1 becasue -32 + 32 is 0
0
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