By making the substitution u=7^x and using part a, derive a hidden quadratic equation
7^(2x-1) +4×7^(x-1) -1/14 =0
and hence find all of the solutions to the above equation.
so, did you do the substitution?
7^(2x) = u^2
the rest is just numbers, so you have a quadratic equation with a bunch of 7's floating around.
I don't understand how you would do that
To derive the hidden quadratic equation, we can use the given substitution u = 7^x. Let's rewrite the equation using this substitution:
7^(2x - 1) + 4 × 7^(x - 1) - 1/14 = 0
Substituting u = 7^x, we have:
7^(2x - 1) = (7^x)^2 = u^2
4 × 7^(x - 1) = 4 × (7^x) × (1/7) = (4/7)u
Now, let's substitute these values back into the equation:
u^2 + (4/7)u - 1/14 = 0
To solve this quadratic equation for u, we can use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 4/7, and c = -1/14. Let's substitute these values into the formula:
u = (-(4/7) ± √((4/7)^2 - 4(1)(-1/14))) / (2(1))
Simplifying further:
u = (-4/7 ± √(16/49 + 4/14)) / 2
u = (-4/7 ± √(16/49 + 8/14)) / 2
u = (-4/7 ± √(16/49 + 24/49)) / 2
u = (-4/7 ± √(40/49)) / 2
u = (-4/7 ± √(40)/√(49)) / 2
Now, let's simplify the square root term:
u = (-4/7 ± 2√10/7) / 2
Divide both the numerator and denominator by 2:
u = (-2/7 ± √10/7)
Since u = 7^x, let's rewrite this in terms of x:
7^x = (-2/7 ± √10/7)
To separate the two solutions, we'll solve for x in each case:
For the positive solution:
7^x = (-2/7 + √10/7)
Taking the logarithm base 7 of both sides:
x = log₇(-2/7 + √10/7)
Similarly, for the negative solution:
7^x = (-2/7 - √10/7)
Taking the logarithm base 7 of both sides again:
x = log₇(-2/7 - √10/7)
Hence, the solutions to the quadratic equation 7^(2x - 1) + 4 × 7^(x - 1) - 1/14 = 0 are:
x = log₇(-2/7 + √10/7)
x = log₇(-2/7 - √10/7)