please help me solve for x:
7^(3-4x)=7^(10x-8)
Thank you!
2007-01-31 14:01:15 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
also please help with:
2^(4x-x^(2))=2^3
2007-01-31 14:02:31 · update #1
In the first equation both bases are 7, so that means the powes must be equal.
3-4x = 10x - 8, is a true statement. From here you solve as you normally would.
Second Equation:
2^(4x-x^(2))=2^3
Again, both bases are two, so you can set the powers equal to each other.
4x - x^2 = 3
Subtract 3 from both sides.
4x - x^2 - 3 = 0
Multiply both sides by negative one, and rewrite in standard form.
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0
x - 3 = 0 or x - 1 = 0
x = 3, or x = 1.
At this point you should go back to the original equation and check both numbers to make sure both are solutions.
2007-01-31 14:09:54 · answer #1 · answered by mirramai 3 · 0⤊ 0⤋
First, get rid of powers and brackets
log(7) (3-4x) = log(7) (10x-8)
or just plain
3-4x = 10x-8
Add 4x, for 3 = 14x -8
Add 8, for 11 = 14x
Divide by 14, for x = 11/14
2007-01-31 22:12:13 · answer #2 · answered by Alan 6 · 0⤊ 0⤋
If the base (7) is the same then the exponents must also be the same. So just solve 3 - 4x = 10x - 8,
and similar for your other one which doesn't show here
2007-01-31 22:10:59 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋