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Identical expressions
- log(2cosx)/log3= four
- logarithm of (2 co sinus of e of x) divide by logarithm of 3 equally 4
- logarithm of (2 co sinus of e of x) divide by logarithm of 3 equally four
- log2cosx/log3=4
- log(2cosx) divide by log3=4
log(2cosx)/log3=4 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
log(2*cos(x))
------------- = 4
log(3)
$$\frac{\log{\left(2 \cos{\left(x \right)} \right)}}{\log{\left(3 \right)}} = 4$$
Detail solution
Given the equation
$$\frac{\log{\left(2 \cos{\left(x \right)} \right)}}{\log{\left(3 \right)}} = 4$$
transform
$$\frac{\log{\left(\frac{2 \cos{\left(x \right)}}{81} \right)}}{\log{\left(3 \right)}} = 0$$
$$\frac{\log{\left(2 \cos{\left(x \right)} \right)}}{\log{\left(3 \right)}} - 4 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
Given the equation
$$\frac{\log{\left(2 w \right)}}{\log{\left(3 \right)}} - 4 = 0$$
$$\frac{\log{\left(2 w \right)}}{\log{\left(3 \right)}} = 4$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(2 w \right)} = 4 \log{\left(3 \right)}$$
This equation is of the form:
By definition log
then
$$2 w = e^{\frac{4}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$2 w = 81$$
$$w = \frac{81}{2}$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$\frac{\log{\left(2 \cos{\left(x \right)} \right)}}{\log{\left(3 \right)}} = 4$$
transform
$$\frac{\log{\left(\frac{2 \cos{\left(x \right)}}{81} \right)}}{\log{\left(3 \right)}} = 0$$
$$\frac{\log{\left(2 \cos{\left(x \right)} \right)}}{\log{\left(3 \right)}} - 4 = 0$$
Do replacement
$$w = \cos{\left(x \right)}$$
Given the equation
$$\frac{\log{\left(2 w \right)}}{\log{\left(3 \right)}} - 4 = 0$$
$$\frac{\log{\left(2 w \right)}}{\log{\left(3 \right)}} = 4$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(2 w \right)} = 4 \log{\left(3 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$2 w = e^{\frac{4}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$2 w = 81$$
$$w = \frac{81}{2}$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
Rapid solution
[src]
x1 = 2*pi - I*im(acos(81/2))
$$x_{1} = 2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}$$
x2 = I*im(acos(81/2)) + re(acos(81/2))
$$x_{2} = \operatorname{re}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}$$
x2 = re(acos(81/2)) + i*im(acos(81/2))
Sum and product of roots
[src]
sum
2*pi - I*im(acos(81/2)) + I*im(acos(81/2)) + re(acos(81/2))
$$\left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right) + \left(\operatorname{re}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right)$$
=
2*pi + re(acos(81/2))
$$\operatorname{re}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)} + 2 \pi$$
product
(2*pi - I*im(acos(81/2)))*(I*im(acos(81/2)) + re(acos(81/2)))
$$\left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right) \left(\operatorname{re}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right)$$
=
(2*pi - I*im(acos(81/2)))*(I*im(acos(81/2)) + re(acos(81/2)))
$$\left(2 \pi - i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right) \left(\operatorname{re}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)} + i \operatorname{im}{\left(\operatorname{acos}{\left(\frac{81}{2} \right)}\right)}\right)$$
(2*pi - i*im(acos(81/2)))*(i*im(acos(81/2)) + re(acos(81/2)))
Numerical answer
[src]
x1 = -150.79644737231 + 4.3942967040245*i
x2 = -56.5486677646163 - 4.3942967040245*i
x3 = -87.9645943005142 + 4.3942967040245*i
x4 = 12.5663706143592 + 4.3942967040245*i
x5 = 56.5486677646163 + 4.3942967040245*i
x6 = -94.2477796076938 + 4.3942967040245*i
x7 = 69.1150383789755 - 4.3942967040245*i
x8 = 3166.72539481851 - 4.3942967040245*i
x9 = 75.398223686155 + 4.3942967040245*i
x10 = 100.530964914873 - 4.3942967040245*i
x11 = 6.28318530717959 - 4.3942967040245*i
x12 = 31.4159265358979 + 4.3942967040245*i
x13 = -43.9822971502571 - 4.3942967040245*i
x14 = -81.6814089933346 + 4.3942967040245*i
x15 = -75.398223686155 - 4.3942967040245*i
x16 = -62.8318530717959 - 4.3942967040245*i
x17 = -37.6991118430775 - 4.3942967040245*i
x18 = 62.8318530717959 + 4.3942967040245*i
x19 = 37.6991118430775 - 4.3942967040245*i
x20 = 50.2654824574367 + 4.3942967040245*i
x21 = -12.5663706143592 + 4.3942967040245*i
x22 = 43.9822971502571 + 4.3942967040245*i
x23 = -87.9645943005142 - 4.3942967040245*i
x24 = -12.5663706143592 - 4.3942967040245*i
x25 = -18.8495559215388 - 4.3942967040245*i
x26 = -6.28318530717959 - 4.3942967040245*i
x27 = 1225.22113490002 + 4.3942967040245*i
x28 = 81.6814089933346 - 4.3942967040245*i
x29 = 100.530964914873 + 4.3942967040245*i
x30 = -62.8318530717959 + 4.3942967040245*i
x31 = -50.2654824574367 + 4.3942967040245*i
x32 = -56.5486677646163 + 4.3942967040245*i
x33 = -100.530964914873 + 4.3942967040245*i
x34 = 7554636800.10092 - 4.3942967040245*i
x35 = 25.1327412287183 - 4.3942967040245*i
x36 = -125.663706143592 + 4.3942967040245*i
x37 = -43.9822971502571 + 4.3942967040245*i
x38 = 87.9645943005142 - 4.3942967040245*i
x39 = -69.1150383789755 - 4.3942967040245*i
x40 = 81.6814089933346 + 4.3942967040245*i
x41 = -6.28318530717959 + 4.3942967040245*i
x42 = 2.9982965206586e-32 - 4.3942967040245*i
x43 = -25.1327412287183 + 4.3942967040245*i
x44 = 18.8495559215388 + 4.3942967040245*i
x45 = 18.8495559215388 - 4.3942967040245*i
x46 = 94.2477796076938 - 4.3942967040245*i
x47 = -50.2654824574367 - 4.3942967040245*i
x48 = 43.9822971502571 - 4.3942967040245*i
x49 = -1514.24765903028 + 4.3942967040245*i
x50 = 87.9645943005142 + 4.3942967040245*i
x51 = -37.6991118430775 + 4.3942967040245*i
x52 = 8.11021064103792e-36 + 4.3942967040245*i
x53 = -94.2477796076938 - 4.3942967040245*i
x54 = 169.646003293849 - 4.3942967040245*i
x55 = 56.5486677646163 - 4.3942967040245*i
x56 = 31.4159265358979 - 4.3942967040245*i
x57 = -31.4159265358979 - 4.3942967040245*i
x58 = 12.5663706143592 - 4.3942967040245*i
x59 = -81.6814089933346 - 4.3942967040245*i
x60 = 69.1150383789755 + 4.3942967040245*i
x61 = 94.2477796076938 + 4.3942967040245*i
x62 = 37.6991118430775 + 4.3942967040245*i
x63 = -18.8495559215388 + 4.3942967040245*i
x64 = 25.1327412287183 + 4.3942967040245*i
x65 = 50.2654824574367 - 4.3942967040245*i
x66 = -31.4159265358979 + 4.3942967040245*i
x67 = -69.1150383789755 + 4.3942967040245*i
x68 = -25.1327412287183 - 4.3942967040245*i
x69 = 62.8318530717959 - 4.3942967040245*i
x70 = 6.28318530717959 + 4.3942967040245*i
x71 = -11755.839709733 + 4.3942967040245*i
x72 = -75.398223686155 + 4.3942967040245*i
x72 = -75.398223686155 + 4.3942967040245*i