Mister Exam
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Similar expressions
- 7^(2x)-6*7^(x)-5=0
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7^(2x)-6*7^(x)+5=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(7^{2 x} - 6 \cdot 7^{x}\right) + 5 = 0$$
or
$$\left(7^{2 x} - 6 \cdot 7^{x}\right) + 5 = 0$$
Do replacement
$$v = 7^{x}$$
we get
$$v^{2} - 6 v + 5 = 0$$
or
$$v^{2} - 6 v + 5 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -6$$
$$c = 5$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 5$$
$$v_{2} = 1$$
do backward replacement
$$7^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(7 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(1 \right)}}{\log{\left(7 \right)}} = 0$$
$$x_{2} = \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}} = \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
$$\left(7^{2 x} - 6 \cdot 7^{x}\right) + 5 = 0$$
or
$$\left(7^{2 x} - 6 \cdot 7^{x}\right) + 5 = 0$$
Do replacement
$$v = 7^{x}$$
we get
$$v^{2} - 6 v + 5 = 0$$
or
$$v^{2} - 6 v + 5 = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -6$$
$$c = 5$$
, then
D = b^2 - 4 * a * c =
(-6)^2 - 4 * (1) * (5) = 16
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 5$$
$$v_{2} = 1$$
do backward replacement
$$7^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(7 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(1 \right)}}{\log{\left(7 \right)}} = 0$$
$$x_{2} = \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}} = \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
Sum and product of roots
[src]
sum
log(5) ------ log(7)
$$\frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
=
log(5) ------ log(7)
$$\frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
product
log(5) 0*------ log(7)
$$0 \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
=
0
$$0$$
0
Rapid solution
[src]
x1 = 0
$$x_{1} = 0$$
log(5)
x2 = ------
log(7)
$$x_{2} = \frac{\log{\left(5 \right)}}{\log{\left(7 \right)}}$$
x2 = log(5)/log(7)