Solve:

Question

(log2 8)^x - (log9 3)^x+1 = 0

Answer 1

^(x+1)
my bad =]

Answer 2

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)

Answer 3

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.

Answer 4

Sorry, did not read the question carefully enough.
Go with MathMate's answer.

Answer 5

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

Answer 6

Quidditch is right.

I did not work with the revised version of the question, and in any case log<sub>93 should have been 1/2.

Work with Quidditch's reponse.</sub>

Answer 7

I agree.
Quidditch is right!!