5^(-x)+7=2x+4 This was on plato

Mathematics

1 answer:

Setler79 [48]3 years ago

3 0

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

You might be interested in

Please please help me on this question

Delvig [45]

-7x^2 - 6x^2 + 3x + 5x - 5 + 2
-13x^2 + 8x - 3

5 0

3 years ago

Please help me with this

Len [333]

Answer:

60 degrees

Step-by-step explanation:

To first solve this problem, we need to figure out the size of an interior angle for a regular hexagon.

This can be done with the formula :

$angle = \frac{(n-2)*180}{n}$ , with n being the number of sides

A hexagon has 6 sides so here is how we would solve for the interior angle:

$\frac{(6-2)*180}{6}=120$ , with n= 6 sides

Now that we know that each interior angle in the hexagon is 120 degrees, we can now turn our attention to the rhombus.

The opposite angles of the rhombus are congruent, so the two larger obtuse angles are congruent, and so are the two smaller acute angles.

It is also important to note that a rhombus is a quadrilateral, so all of its interior angles add up to 360 degrees.

Looking at the rhombus, we already know one of the angles because it is shared by the interior angle of the hexagon, so the two larger angles in the rhombus are both 120 degrees.

But what about the smaller angles? All we need to do is subtract the two larger angles form 360 and divide by 2 to find the angle.

$\frac{360-2(120)}{2} = 60$ , so the smaller angle in the rhombus is 60 degrees.