## Solution of Hard Algebra Puzzles by only Basic Exponent Concepts and Mathematical Reasoning

Solve for $x$ in $(x^2-5x+5)^{(x^2-11x+30)}=1$.

Looks hard?

Rest assured, you too can solve it. Just use basic exponent concepts and a bit of logical reasoning.

The two puzzles below are both of same type. So if you solve the first one, the second shouldn't take much time at all.

That's why for solving the second puzzle, we have given much less time than the first.

### The Exponent Algebra Puzzles

**Exponent Algebra ****Puzzle 1.** Solve for $x$ in $(x^2-7x+11)^{(x^2-13x+42)}=1$. Time to solve 10 minutes.

**Exponent Algebra ****Puzzle 2.** Solve for $x$ in $(x^2-5x+5)^{(x^2-11x+30)}=1$. Time to solve 5 minutes.

**Hint.** No hard maths needed. Simple concepts and systematic logical thinking only.

As you are trying to solve the puzzles, let's say a few words before we explain the detailed steps to the solutions.

If you want **to skip the preamble** and move straight to the solution, **click here.**

### Preamble

If you are preparing for tough competitive exams, these two problems shouldn't take much time for you to solve. But you know, most people out there are not preparing for any tough competitive exams.

The knowledge needed to solve these puzzle quickly, may be in a few minutes, are so very basic that all of you reading this should be able to solve these easily.

But knowing the concepts only are not enough, you would also need focused logical thinking.

Usually we were taught how to solve math problems using math theorems or formulas step by routine steps, and **not by reasoned logical thinking.**

In contrast, these two puzzles perfectly **highlight the systematic logical approach to math problem solving.**

Let's get down to solving the puzzles.

### Solution to exponent algebra puzzle 1 by basic concepts and mathematical reasoning

**Exponent **** Algebra Puzzle 1:** Solve for $x$ in $(x^2-7x+11)^{(x^2-13x+42)}=1$.

**First question** that automatically pops up in our head,

In

what situationswould the value of aBasealgebraic expression with anExponentbecome 1?

This is the **most important question in this puzzle.** If you get **answers to this question fully,** your puzzle is solved immediately. This is the **critical question** in the puzzle.

It is important to

forgetthat the given equation is a quadratic polynomial to the power of a second quadratic polynomial, a form that perhaps you have never seen. Just get out of this surprise shock, if any, and imagine the equation to be simplya variable to the power a second variable.

Convert in your mind the original expression to an **abstracted** form,

$(x^2-7x+11)^{(x^2-13x+42)}=1$

$\Rightarrow \text{variable1}^{\text{variable2}}=1$.

#### Asking the Critical Question while Solving a Problem and Thinking Exhaustively for all possible Answers

This is what we call **THE Critical Question.** There is only one in this puzzle. In a complex problem there may be a series of such questions. As you get answer to one, a new question comes up.

To solve a complex problem,

you must find the critical questionthat stands between you and the solution, and then find the right answer to the question.

**Secondly,** the importance of **ANSWERING FULLY** can never be overemphasized.

This is what we call **Exhaustive Thinking.**

Assume for a moment that you are sitting in front of me in your home and I ask you the question,

How many ways can you go to your bathroom from where you are sitting?

You may look at me thinking, "this man must have gone crazy."

Or if you are in a mood to be very thorough, you may find many answers by first moving back and forth all through your house and then finally to the bathroom.

But no need to count. It is enough to get an idea of thoroughness in this **open-ended** question with uncountable number of answers.

Instead, suppose I ask you,

In how many unique ways can you arrange the letters a, b, c and d? Don't use any formula.

You now face a closed set of specific number of arrangement possibilities.

**Answer will be a specific number** and you have to **arrange and rearrange the letters systematically** to know yourself that you have finally arrived at the correct answer.

This needs **exhaustive and systematic thinking for enumerating all possible arrangements.**

*Ability to think exhaustively is valuable.* The secret is, *we are generally not habituated to think exhaustively.*

Okay, enough talk, let's solve the puzzle.

#### The first answer to the critical question

*First basic exponent concept,*

If the

exponent is zero,thevalue of the expression with the exponent becomes 1.

We know it from **$a^0=1$**, whatever be the value of the Base $a$.

So $x^2-13x+42$ being the exponent, $x^2-13x+42=0$ will give us our first set of two values.

Factorizing the quadratic polynomial on the left of the equation is easy and you will get your first two valid values,

$x^2-13x+42=0$,

Or, $(x-6)(x-7)=0$.

For $x=6$ and $x=7$ the quadratic equation is satisfied, the exponent expression becomes zero, and our given puzzle equation value becomes 1.

**Any more possibilities?**

As you ask yourself this question, you remember the second commonly known possibility from $(1)^n=1$.

This is the second basic exponent concept,

If the

Base value is 1, whatever be its real Exponent value, the value of theexpression remains at 1.

#### The second answer to the critical question

When the **Base expression is 1**, whatever be the exponent, the **value of the equation with exponent remains at 1.**

Let's quickly pick up the two more valid answers from,

$x^2-7x+11=1$,

Or, $x^2-7x+10=0$,

Or, $(x-2)(x-5)=0$.

Two more answers in your bag are, 2 and 5.

Till now the answers are, 2, 5, 6, and 7.

**Is that all?**

This is an **important point in the whole solution process.** You may stop here without finding any other clue. Nothing wrong, majority stop here with 4 valid values of $x$.

But if you still ask yourself, "Any more possibilities?" a few times, chances are you will become an exhaustive thinker. Suddenly a new possibility will pop up in your head.

Isn't it a fact that, $(-1)^n=1$, where $n$ is an even integer?

Of course it is true.

#### Third possible answer to the critical question

Basic derived concept,

If

Base expression$x^2-7x+11=-1$andexponent$x^2-13x+42$ isevenfor the values of $x$ for which the Base expression is $-1$, then, $(x^2-7x+11)^{(x^2-13x+42)}=1$.

This is **dependent value situation** where **primary condition is the Base value of $-1$.**

**Only if this condition is true,** we'll look for values of the exponent with which base to the power exponent becomes 1.

This is how you get the two *bonus answer values* from,

$x^2-7x+11=-1$,

Or, $x^2-7x+12=0$,

Or, $(x-3)(x-4)=0$.

For $x=3$ and $x=4$ the Base expression value becomes $-1$.

What about **values of the exponent expression for $x=3$ and $x=4$?**

For $x=3$,

$x^2-13x+42=9-39+42=12$.

Thank heavens, it is even and gives us our 5th answer.

And for $x=4$?

For $x=4$,

$x^2-13x+42=16-52+42=6$. Still better. This one too.

Together till now we have then **six values of $x$** that are all correct answers,

2, 3, 4, 5, 6 and 7.

**Any more possibilities?**

You never know, a new answer may pop up. After all, **instead of a system**, we are **trying to ensure exhaustivity from our basic concepts in math**, and we are not mathematicians.

As far as we know, 2, 3, 4, 5, 6 and 7 are the set of all valid values of $x$ and are the answers to the puzzle.

### Solution to exponent algebra puzzle 2 by basic concepts and mathematical reasoning

**Exponent **** Algebra Puzzle 2:** Solve for $x$ in $(x^2-5x+5)^{(x^2-11x+30)}=1$.

Knowing the anatomy and geography of such a problem by now we waste no time and get our first pair of valid values from,

$(x^2-11x+30)=(x-5)(x-6)=0$.

5 and 6 are the first two valid values of $x$.

Then the second pair from,

$(x^2-5x+5)=1$

Or, $x^2-5x+4=(x-1)(x-4)=0$.

1 and 4 are the second pair of valid values.

Finally we try to get the third pair of valid values.

Why "try"? Because of dependent value situation, the value of $x$ that makes the Base value $-1$ may not make the exponent expression value even. That's why.

We have to try and see.

$(x^2-5x+5)=-1$

Or, $x^2-5x+6=(x-2)(x-3)=0$.

For $x=2$ exponent expression value is,

$(x^2-11x+30)=4-22+30=12$. It is **even.** $x=2$ is our 5th answer value.

For $x=3$ exponent expression value is,

$(x^2-11x+30)=9-33+30=6$. It is **even.** $x=3$ is then our 6th answer value.

We have then 6 valid values of $x$ as answers,

1, 2, 3, 4, 5 and 6.

The puzzles are not really hard, isn't it?

### Last one

Will you try to form a new similar puzzle?

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