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The answer is correct but is it possible that you could go into further detail on how you ended up with your answers?

I AGREE WITH ALAN:

Solve for x over the real numbers:
2^(2 x + 19) = 2^((x - 1)/2)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2) (2 x + 19) = 1/2 log(2) (x - 1)

Divide both sides by log(2):
2 x + 19 = (x - 1)/2

Expand out terms of the right hand side:
2 x + 19 = x/2 - 1/2

Subtract x/2 + 19 from both sides:
(3 x)/2 = -39/2

Multiply both sides by 2/3:
 x = -13

I get the following:

As follows:

The answer is clearly not -9:

deleted

Oh, the \(LaTeX\) version is much clearer.

Simply both sides first so you can isolate the variable.

Remember, isolate, isolate, isolate.

\(x=-1,0,1\)

\(x\) can be -1, 0, or 1.

x = 1/x + 8     is what is being asked   multiply by x

x^2 = 1+8x     re-arrange

x^2-8x-1=0     use quadratic formula   a = 1 b = -8  c = -1

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

{8+-sqrt(64+4(1)(1))   } / 2

4 +- sqrt(17*4)/2

4+- 2(sqrt17)/2

4 +- sqrt 17

Thanks answer guest.   

Asker guest.      Can you tell me what the equation of the red line is?

√5x^2 - (8√5 +10)x + (20√5 +40) = 0

\(\sqrt{5}\cdot x^2-(8\cdot \sqrt{5}+10)\cdot x+(20\cdot \sqrt{5}+40)=0\\ \sqrt{5}=w\\ wx^2-(8w+10)\cdot x+(20w+40)=0\)

a                   b                            c

\(x = {8w+10 \pm \sqrt{(8w+10)^2-4(w)(20w+40)} \over 2w}\\ x=\frac{8w+10\pm\sqrt{64w^2+160w+100-80w^2-160w}}{2w}\\ x=\frac{8w+10\pm\sqrt{100-16w^2}}{2w}\\ x=\frac{8w+10\pm\sqrt{(10+4w)\cdot(10-4w)}}{2w}\\ \ w=\sqrt{5}\\ x=\frac{27,88854\pm\sqrt{20}}{4,472136}\)

\(x_1=7,23607\\ x_2=5,23607\)

  !

Solve for x:
8 + 1/x = x

Bring 8 + 1/x together using the common denominator x:
(8 x + 1)/x = x

Multiply both sides by x:
8 x + 1 = x^2

Subtract x^2 from both sides:
-x^2 + 8 x + 1 = 0

Multiply both sides by -1:
x^2 - 8 x - 1 = 0

Add 1 to both sides:
x^2 - 8 x = 1

Add 16 to both sides:
x^2 - 8 x + 16 = 17

Write the left hand side as a square:
(x - 4)^2 = 17

Take the square root of both sides:
x - 4 = sqrt(17) or x - 4 = -sqrt(17)

Add 4 to both sides:
x = 4 + sqrt(17) or x - 4 = -sqrt(17)

Add 4 to both sides:
 
 x = 4 + sqrt(17)         or                x = 4 - sqrt(17)

https://web2.0calc.com/questions/find-a-linear-inequality-with-the-following-solution_1

Thanks

Two points o the line are   3,1    and  1,-2        slope then equals    (y1-y2)/(x1-x2) = (1- -2) / (3-1) = 3/2

y = 3/2 x   +  b    find b by plugging in one of the points

1 = 3/2 (3) + b     b = -7/2

so the line equation is     y = 3/2 x - 7/2

The yellow shaded area is thus       y> 3/2 x - 7/2      Multiply through by 2

                                                        2y > 3x - 7           re-arrange

                                                        2y - 3x + 7 >0

This sequence consists of Fibonacci Numbers raised to the power of 2 as follows:

Fibonacci Sequence: 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55.......[10 terms]

Your sequence        : 1, 2, 2, 4, 8, 32, 256........... Your 7th term = 2^(Fib 6th term) = 2^8 = 256

Therefore: Your 35th term =2^34F. 34th Fib. term =5,702,887  =2^5,702,887, which will give 8 as the last digit as can be seen here in the last 20 digits:.....49292157765605654528.

put it into the latex thing

OHHHH FANCY

\(\color{red}{jimin}_{\color{orange}{\LaTeX}_{\color{green}{\LaTeX}}{\color{blue}{\LaTeX}}}^{\color{orange}{\LaTeX}^{\color{green}{\LaTeX}}{\color{blue}{\LaTeX}}}\color{purple}{\LaTeX} \)

I just wrote it out, and got 2

If (x, y) is a point on the graph its distance from the origin will be sqrt(x^2 + y^2),

and it's this you have to find the minimum value of.

However, since x^2 + y^2 will get you the same point it's easier if you use this.

Use the equation of the curve to substitute for y^2. 

The last digit of a multiplication problem is the last digit of the product of the two unit digits.

First make a list of the unit digits of the sequence and see if there is a pattern.

1, [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2]

It looks like after 1 there is a 6 digit repeating pattern [2, 4, 8, 2, 6, 2]

Ignoring the 1 at the beginning of the sequence the problem is now find the 34th term in the pattern that was found.

The largest multiple of 6 under 34 is 30 so the 30th term will be the last 2 in the pattern.

30- 2

31- 2

32- 4

33- 8

34- 2

The unit digit of the 35th term in the original sequence is \(\boxed{2}\)