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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: fredrick on November 15, 2007, 02:06:01 pm

Title: im stumped
Post by: fredrick on November 15, 2007, 02:06:01 pm

4^x=10-4^(x+1)

solve for x :shock:

Title: im stumped
Post by: brendan on November 15, 2007, 02:07:47 pm

hint: let a = 4^x

Title: im stumped
Post by: fredrick on November 15, 2007, 02:10:09 pm

a=10-4^(x+1)??

Title: im stumped
Post by: fredrick on November 15, 2007, 02:24:23 pm

ok i got the right answer but is there any bette way of doing it? here how i went:
4^x=10-4^(x+1)
xlog(4)=log(10-4^(x+1))
xlog(4)=1/(log(4)^(x+1))
xlog(4)=log(4)^(1-x)
xlog(4)=(1-x)log(4)
xlog(4)=log(4)-xlog(4)
2xlog(4)=log(4)
x=1/2

Title: im stumped
Post by: Toothpaste on November 15, 2007, 02:26:57 pm

4^x = 10 - 4^x  * 4

let a = 4^x

a= 10 - 4a

5a = 10

a = 2



=> 2 = 4^x

2 = 2^(2x)

1 = 2x

x = 1/2

Title: im stumped
Post by: fredrick on November 15, 2007, 02:33:14 pm

Quote from: "Toothpick"

4^x = 10 - 4^x  * 4

let a = 4^x

a= 10 - 4a

5a = 10

a = 2



=> 2 = 4^x

2 = 2^(2x)

1 = 2x

x = 1/2

ok thats much easier, but i neva knew 4^(x+1)=4(4^x) maybe cause this is unit 1+2 methods??

Title: im stumped
Post by: Toothpaste on November 15, 2007, 02:36:52 pm

Quote from: "fredrick"

ok thats much easier, but i neva knew 4^(x+1)=4(4^x) maybe cause this is unit 1+2 methods??

Um, you're supposed to know it in 1&2.

Log laws/ exp. laws.

a^(p+q) = a^p  *  a^q


for the question you asked:
4^(x+1) = 4^1  *  4^x

Title: im stumped
Post by: AppleXY on November 15, 2007, 03:18:07 pm

Exactly. Lol, you should've learnt your log rules way back in 1/2 :p

Title: im stumped
Post by: Collin Li on November 15, 2007, 06:26:18 pm

Those indicial rules should have been learnt in year 7 and 8, haha. Just need to revive a bit of that memory.