Solve for x:

8^x + 2^x = 30

As usual, watch the video for a solution.

Trick You Should Know For Harvard Entrance Exam Problem

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Trick You Should Know For Harvard Entrance Exam Problem

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Using exponent rules we have:

8^x + 2^x = 30
(2³)^x + 2^x = 30
(2^3x) + 2^x = 30
(2^x)³ + 2^x = 30

Now let’s substitute y = 2^x. Thus we have a cubic equation:

y³ + y = 30
y³ + y – 30 = 0

If we can figure out a root of this equation, we can reduce it to the product of a linear equation and a quadratic equation, both of which we can readily solve. By the rational root theorem, we have candidate rational roots are plus or minus factors of the constant term 30 divided by the factors of the leading coefficient of 1. This gives candidates of ± 1, 2, 3, 5, 6, 10, 15, 30. Testing y = -1 we get:

-1 – 1 – 30 = -32

Any larger negative value will just be more negative. Testing y = 1, 2, 3 we get:

1 + 1 – 30 = -28
8 + 2 – 30 = -20
27 + 3 – 30 = 0

So y = 3 is a root, and it is the only rational root of this equation.

Therefore y³ + y – 30 is divisible by y – 3. Performing long division we get:

This means we have:

(y – 3)(y² + 3y + 10) = y³ + y – 30

Setting the equation equal to 0, we have the linear factor or the quadratic factor has to be equal to 0, giving:

y – 3 = 0
y = 3

y² + 3y + 10 = 0
(using quadratic formula)
y = -1.5 ± 0.5 i√31

We previously made the substituton:

y = 2^x
ln y = ln 2^x
ln y = x ln 2
x = (ln y)/(ln 2)

Thus we have 3 possible solutions:

y = 3
x = (ln 3)/(ln 2)

y² + 3y + 10 = 0
x = (ln (-1.5 ± 0.5 i√31))/(ln 2)

It is important to check for extraneous solutions, so it is good to test these values to make sure they do solve the original equation.

You can check in WolframAlpha, for example:

x = (ln 3)/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+3%2Fln+2%29%2B2%5E%28ln+3%2Fln+2%29

x = (ln (-1.5 + 0.5 i√31))/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29%2B2%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29

x = (ln (-1.5 – 0.5 i√31))/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29%2B2%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29

Thus we have found 3 solutions to the original equation. What a fun problem!

Anna says the number 2021 is fantabulous. All such numbers obey a property.

If any element of the set {m, 2m + 1, 3m} is fantabulous for a positive integer m, then all elements of the set are fantabulous.

Is the number 2021²⁰²¹ fantabulous?

As usual, watch the video for a solution.

Fantabulous Numbers

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To Fantabulous Numbers

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

A mathematician would likely think all positive integers are fantastic and fabulous, so it would stand to reason that all numbers are fantabulous. Can we prove it?

Since m = 2021 is fantabulous, we can readily calculate other numbers

2m + 1 = 4043
3m = 6063

But now let’s make a set {n, 2n + 1, 3n} with 2n + 1 = 6063. We then have:

n = 3031

Iterating with 2p + 1 = 3031, we have:

p = 1515

Taking 1515 = 3q, we have

q = 505

Take this as the form 2r + 1 to get

r = 252

We can iterate in this fashion to find fantabulous numbers

252/3 = 84
84/3 = 28
2(28) + 1 = 57
57/3 = 19
(19 – 1)/2 = 9
9/3 = 3
3/3 = 1

So 2021 being fantabulous implies 1 is.

We already know m being fantabulous implies 2m + 1 is. If we could also show it implies 2m is fantabulous, then we would have 1 being fantabulous implies 2, 3 are, and then iterating we would have 4, 5, 6, 7 are, and so on, so that all positive integers are fantabulous, including 2021²⁰²¹.

We can proceed like in the last calculation.

Let m be fantabulous. Then we will have fantabulous numbers from the operations 2m + 1, 3m. If m is odd we can perform (m – 1)/2, and if m is a multiple of 3 we can do m/3. So the fantabulous numbers are:

m
3m
2(3m) + 1 = 6m + 1
2(6m + 1) + 1 = 12m + 3
(12m + 3)/3 = 4m + 1
((4m + 1) – 1)/2 = 2m

Thus m being fantabulous implies 2m is. Along with m implies 2m + 1 and 1 being fantabulous, we have that all positive integers are fantabulous, including 2021²⁰²¹.

Reference

2021 European Girls’ Mathematical Olympiad (EGMO)
https://www.egmo.org/egmos/egmo10/solutions.pdf
https://www.egmo.org/egmos/egmo10/

Here’s a fun little puzzle. What’s the length of the top side?

When I shared this puzzle, most people were able to find the correct answer. You might say that 99 percent could find the answer.

But let’s make one change to the puzzle, which I credit to a comment by Michael Timothy. Change the left length from 4 to 3, and now see if people can solve for the top side.

So what do you think? What is the value of x in this shape? Incredibly many people fail to get the correct answer. So what is the correct answer?

As usual, watch the video for a solution.

What Is The Length? 99% Fail

Or keep reading.

.
.
.
.
M
I
N
D
.
Y
O
U
R
.
D
E
C
I
S
I
O
N
S
.
P
U
Z
Z
L
E
.
.
.
.
Answer To

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Let’s tackle the first problem.

So how do many people solve this? They construct the hypotenuse of the 3-4 right triangle, which will have a length of 5. And then they will construct a vertical line, saying it has a distance of 3. So they get another 3-4-5 right triangle, and a shape like this:

If this figure were accurate, then the value of the top side would be 5 + 3 = 8. And this is the correct answer!

However, there are a couple of unjustified steps in this derivation. You would need to show the horizontal 5 is perpendicular to the vertical line, and you would also need to show the length of 3 reaches the endpoint.

So what’s the danger of using this method, if the answer was correct anyway?

The problem is this derivation will not work in general. Let’s see what happens when we use it to solve problem 2.

By this same method, the top side should be 5 + 4 = 9. But this is not the correct answer!

The construction makes a couple of mistakes: the blue line at the bottom is not parallel to the top line, and the vertical line marked 3 is not actually a distance of 3.

The point is made clear when the diagram is re-drawn to scale.

I find it interesting that students are basing their reasoning on the diagrams themselves.

In my math education, we were always taught never to assume diagrams are drawn to scale, and we were also taught not to assume angles like perpendicular or parallel lines.

Teachers are not being sinister by presenting diagrams not drawn to scale. Diagrams are meant to show relations in the big picture. In real life many diagrams are not drawn to scale: think about schematic maps of subway lines, atomic models, solar models, or maps of countries of the world. You should not assume a diagram is drawn to scale!

Solution to problem 2

My technique to solve this problem is to consider the distance from the bottom left corner to the top right corner.

This works out to √(4² + (3 + 5)²) = √80, as it is the same distance of the hypotenuse of a right triangle with legs of 4 and 8.

Now go back to the original figure.

And then consider just the right triangle as shown.

We can then readily solve for x using the distance formula:

x² = (√80)² – 3²
x² = 71

Since the side length is a positive number, we solve that x = √71.

In geometry you cannot always trust the diagrams, so the problem is a little bit trickier than it seems at first!

Source

Comment from Michael Timothy
https://www.youtube.com/watch?v=amLDMD8KPzQ&lc=Ugw4bQRoE0tit8sLNPR4AaABAg.8xArwe3DbZu8xCMU8Idv0-
Presh, please redo this problem using the bad solution given below, based on 3-4-5 triangles. Then do that proof again, except make the far left 4 a 3. That solution gives the answer ‘9’, while your correct method here gives root(71).

solar system nasa not scale size
https://commons.wikimedia.org/wiki/File:Solar_System_graphic_by_NASA.png
https://en.wikipedia.org/wiki/File:Scale_of_the_Solar_System.png

original video
https://www.youtube.com/watch?v=kfijuP5HDjU

Jakub Nowosad, CC BY-SA 4.0 , via Wikimedia Commons
https://en.wikipedia.org/wiki/Harry_Beck
https://en.wikipedia.org/wiki/Harry_Beck#/media/File:Tube_map_1908-2.jpg
https://commons.wikimedia.org/wiki/File:Harry_Beck%27s_tube_map_at_Finchley_central_tube_station_-_geograph.org.uk_-_6206154.jpg
Harry Beck’s tube map at Finchley central tube station by Mike Quinn, CC BY-SA 2.0 , via Wikimedia Commons