Original post by judogainz
2^x+4^x+1=8
How do you solve this?
How do you solve this?
Having everything with a base of 2^... may help.
Original post by SeanFM
Having everything with a base of 2^... may help.
taht was what is was thinking how would i get the other one to 2?
Original post by judogainz
taht was what is was thinking how would i get the other one to 2?
Law of indices.
How can 4^x be written in terms of 2^...? (Hint: 4=2^2)
Original post by SeanFM
Law of indices.
How can 4^x be written in terms of 2^...? (Hint: 4=2^2)
How can 4^x be written in terms of 2^...? (Hint: 4=2^2)
i know that you square it, but would it go to 2^x+3^2(x+1) ?
Original post by judogainz
i know that you square it, but would it go to 2^x+3^2(x+1) ?
The 3 should be a 2, but yes, well done
Original post by SeanFM
The 3 should be a 2, but yes, well done
so you ignore the base and just use the pwoers
Im left with x+2(x+1)=8
do i just solve it from here?
Original post by judogainz
so you ignore the base and just use the pwoers
Im left with x+2(x+1)=8
do i just solve it from here?
Im left with x+2(x+1)=8
do i just solve it from here?
You can't quite jump like that
So the things with x in are now of the form 2^...
Giving you 2^x + 2^(x+2) - 8 = 0, and you know that 2^(x+2) = 2^x * 2^2 by law of indices. After you've changed that, can you think of how to solve it? (Hint: a substitution turns it into a quadratic).
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