Solve:
(log2 8)^x - (log9 3)^x+1 = 0
^(x+1)
my bad =]
Isn't log28 = 3 ?
so your equation becomes
3^x - 3^(x+1) = 0
3^x(1 - 3) = 0
-2(3^x) = 0
3^x = 0
no real numbers satisfies that equation.
(3 raised to what number equals zero ??
try x = -100, then 3^-100 = 1.94 x 10^-48 , pretty small and close to zero, but not zero.
x would have to be negative infinity, but that is not a real number)
Note that
log28 = 3
and
log93 = 1/3
Substitute
y=3^x, then
y - 1/y +1 =0
which transforms to a quadratic equation for which y can be solved.
Since a^x > 0 for a>0 and x∈ℝ,
one of the solutions where y<0 has to be rejected.
Continuing with the other solution, I get
y = 3^x = (-1+sqrt(5))/2
By taking ln on both sides, I get
x=ln(y)/ln(3)
=-0.44 approx.
Sorry, did not read the question carefully enough.
Go with MathMate's answer.
log2(8)=3
log9(3)=0.5
So,
(3)^x = (0.5)^(x+1)
ln of both sides
(x)(ln(3) = (x+1)(ln(0.5)
(ln(3) - ln(0.5))(x)= ln(0.5)
calculate ln(3) and ln(0.5)
solve for x
Quidditch is right.
I did not work with the revised version of the question, and in any case log93 should have been 1/2.
Work with Quidditch's reponse.
I agree.
Quidditch is right!!