Mister Exam
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- Equation:
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Equation x^3=125
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- Equation (2log9)^2-3log9x+1=0
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Identical expressions
- (two log9)^2-3log9x+ one = zero
- (2 logarithm of 9) squared minus 3 logarithm of 9x plus 1 equally 0
- (two logarithm of 9) squared minus 3 logarithm of 9x plus one equally zero
- (2log9)2-3log9x+1=0
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- (2log9)²-3log9x+1=0
- (2log9) to the power of 2-3log9x+1=0
- 2log9^2-3log9x+1=0
- (2log9)^2-3log9x+1=O
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Similar expressions
- (2log9)^2-3log9x-1=0
- (2log9)^2+3log9x+1=0
(2log9)^2-3log9x+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 (2*log(9)) - 3*log(9*x) + 1 = 0
$$\left(- 3 \log{\left(9 x \right)} + \left(2 \log{\left(9 \right)}\right)^{2}\right) + 1 = 0$$
Detail solution
Given the equation
$$\left(- 3 \log{\left(9 x \right)} + \left(2 \log{\left(9 \right)}\right)^{2}\right) + 1 = 0$$
$$- 3 \log{\left(9 x \right)} = - 4 \log{\left(9 \right)}^{2} - 1$$
Let's divide both parts of the equation by the multiplier of log =-3
$$\log{\left(9 x \right)} = \frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}$$
This equation is of the form:
By definition log
then
$$9 x = e^{\frac{- 4 \log{\left(9 \right)}^{2} - 1}{-3}}$$
simplify
$$9 x = e^{\frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}}$$
$$x = \frac{e^{\frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}}}{9}$$
$$\left(- 3 \log{\left(9 x \right)} + \left(2 \log{\left(9 \right)}\right)^{2}\right) + 1 = 0$$
$$- 3 \log{\left(9 x \right)} = - 4 \log{\left(9 \right)}^{2} - 1$$
Let's divide both parts of the equation by the multiplier of log =-3
$$\log{\left(9 x \right)} = \frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$9 x = e^{\frac{- 4 \log{\left(9 \right)}^{2} - 1}{-3}}$$
simplify
$$9 x = e^{\frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}}$$
$$x = \frac{e^{\frac{1}{3} + \frac{4 \log{\left(9 \right)}^{2}}{3}}}{9}$$
Rapid solution
[src]
2
1 16*log (3)
- + ----------
3 3
e
x1 = ---------------
9
$$x_{1} = \frac{e^{\frac{1}{3} + \frac{16 \log{\left(3 \right)}^{2}}{3}}}{9}$$
x1 = exp(1/3 + 16*log(3)^2/3)/9
Sum and product of roots
[src]
sum
2
1 16*log (3)
- + ----------
3 3
e
---------------
9
$$\frac{e^{\frac{1}{3} + \frac{16 \log{\left(3 \right)}^{2}}{3}}}{9}$$
=
2
1 16*log (3)
- + ----------
3 3
e
---------------
9
$$\frac{e^{\frac{1}{3} + \frac{16 \log{\left(3 \right)}^{2}}{3}}}{9}$$
product
2
1 16*log (3)
- + ----------
3 3
e
---------------
9
$$\frac{e^{\frac{1}{3} + \frac{16 \log{\left(3 \right)}^{2}}{3}}}{9}$$
=
2
1 16*log (3)
- + ----------
3 3
e
---------------
9
$$\frac{e^{\frac{1}{3} + \frac{16 \log{\left(3 \right)}^{2}}{3}}}{9}$$
exp(1/3 + 16*log(3)^2/3)/9