Thread: Functions and logs

Hey everyone, I need some help with my homework. I've been trying since Monday evening to solve them but I think my brain is dry already... I already solved the other five but these three just won't budge. Thanks!

1) Describe each eq'n in the ff. system. Sketch the graphs of the eq'ns in the system using only one set of axes.

{ y=-3/x
{
{ y=x^2-3x-1

2) Given that r1, r2, and r3 are the roots of the 3rd deg. polynomial function
P(x) = x^3-2x^2-23x+k, find the exact value of the squares of the roots r1^2+r3^2+r3^2.

3) Solve for x: 27(4^x)+19(6^x)-8(9^x)=0

Thanks guys!

Hello, umbraculum!

For #2, you need some "equation theory" . . .

2) Given that a, b, and c are the roots of the cubic: .P(x) .= .x³ - 2x² - 23x + k,
. . find the exact value of: .a² + b² + c².

The theory applies to all polynomials, but I'll explain it for this cubic only.

Given a cubic equation: .x³ + px² + qx + r .= .0 .with roots a, b, c:
. . a + b + c .= .-p, . ab + bc + ac .= .q, . abc .= .-r


We have: .(1) a + b + c .= .2, . (2) ab + bc + ac .= .-23, . (3) abc .= .-k


Square equation (1): .(a + b + c)² .= .2²

And we have: .a² + b² + c² + 2ab + 2bc + 2ac .= .4

Factor: .a² + b² + c² + 2(ab + bc + ac) .= .4
. . . . . . . . . . . . . . . . . . \_________/
. . . . . . . . . . . . . . . . . . . .This is -23


So we have: .a² + b² + c² + 2(-23) .= .4 . → . a² + b² + c² .= .50

Just to add to Soroban's solution, he used a simple (but useful) theorem all the way from the Renassiance called "Viete's Theorem".

that was amazing soroban, did not know that

Hellom umbraculum!

#3 is a freaky one . . .

3) Solve for x: .27(4^x) + 19(6^x) - 8(9^x) .= .0

Note that: .4^x .= .(2²)^x .= .2^{2x} .= .(2^x)²

. . . . . . . . 6^x .= .(2·3)^x .= .(2^x)(3^x)

. . . . . . . .9^x .= .(3²)^x .= .3^{2x} .= .(3^x)²


The equation becomes: .27(2^x)² + 19(2^x)(3^x) - 8(3^x)² .= .0


Let: .a = 2^x, .b = 3^x

And we have: .27a² + 19ab - 8b² .= .0

. . which factors: .(a + b)(27a - 8b) .= .0

Hence: .(1) a + b .= .0 . or . (2) 27a - 8b .= .0


(1) a + b .= .0 . → . 2^x + 3^x .= .0 . → . 2^x .= .-3^x

But this is impossible.
2^x and 3^x are both positive; one cannot be the negative of the other.


(2) 27a - 8b .= .0 . → . 27(2^x) - 8(3^x) .= .0

We have: .3³(2^x) .= .2³(3^x) . → . (2^x)/(3^x) .= .(2³)/(3³)

Hence: .(2/3)^x .= .(2/3)³

Therefore: . x = 3