Please help me solving this sum?

Question

9^x+2 - 6 X 3^x+1 +1=0 I've got to find the value of x (variable).

Answer 1

9^x-6X3^x+4=0
3^2x-6X3^x+4=0
put 3^x=y,
y^2-6y+4=0
find roots by -b+or-rootoverb^2-4ac/2a a,b,c ax^2+bx+c=0
you get imaginary roots.

Answer 2

Not easy to figure out just what the problem is. Some parentheses would've helped. Also, I guess your big X means multiply. Let me try to rewrite and do it ...

[9^(x+2) - 6] [3^(x+1) + 1] = 0

This is already factored into two binomials, so just set each one equal to zero:

9^(x+2) = 6
(3^2)^(x+2) = 6
3^(2x+4) = 6

Take the log of both sides:

(2x+4) ln 3 = ln 6 = ln 3 + ln 2
2x + 4 = (ln 3 + ln 2)/ln 3 = 1 + (ln 2)/(ln 3)
2x = (ln 2)/(ln 3) - 3
x = (ln 2)/(2 ln 3) - 3/2 = [(ln 2)/(ln 3) - 3]/2 = -1.1845351

The above line is the answer, and the solution checks. Now do the same for the other binomial:

3^(x+1) = -1
(x+1) ln 3 = [undefined]

You can't take the log of a negative number, so you only have one solution for x.

Answer 3

Because (3^x)^2=3^(2x)=(3^2)^x, this is a quadratic equation in the variable 3^x.

Rewrite it:

(3^x)^2- 6 * 3^x+2=0

Now use the quadratic formula to solve for 3^x:

3^x=(6 \pm \sqrt{36-8})/2=
(6 \pm 2 \sqrt{7})/2=
3 \pm \sqrt{7}

Now take the natural logarithm of both sides

x ln(3=ln(3^x))=ln(3 \pm \sqrt{7})

and divide by ln(3):

x=ln(3 \pm \sqrt{7})/ln(3)


Note that both 3 \pm \sqrt{7} are positive so ln(3 \pm \sqrt{7}) is defined.

Answer 4

9^x=3^2x
3^2x+2-6x3^x+1+1=0
(3^2x -6x3^x +1) 2+1=0
(3^x-1)^2=-3 so, you cann't take the root square of(-3)

so,there is no value for x

Answer 5

I don't know whether you have the equation correctly written here or not, but the principle of solving it is to convert it into quadratic equation:
9^x-6*3^x+4=0
3^2x-6*3^x+4=0
3^x=a
a^2-6a+4=0
D=36-16=20
a=(6+/-SQRT20)/2
a(1)=(6+4.47)/2=5.235
a(2)=(6-4.47)/2=0.764
3^x(1)=5.235 x(1)=log(3) 5.235=1.507
3^x(2)=0.764 x(2)=log(3) 0.764=-0.245