Solutions to an Exponential

Denote $A = x^2 - 17x + 71$ and $B = x^2 - 34x + 240$ . Since we seek integral $x$ , $A$ and $B$ are integral. We proceed by messy casework.

Case 1 : $A = 1$ , $B$ is anything.

Solving our case conditions gives: $1 = A = x^2 - 17x + 71$

$\Rightarrow (x - 7)(x - 10) = 0$

and so our solutions are $x = 7$ or $10$ .

Case 2 : $A ≠ 0, B = 0$ .

$0 = B = x^2 - 34x + 240$

$\Rightarrow (x - 24)(x - 10) = 0$

If $x = 24, A = 239 ≠ 0$ .

If $x = 10, A = 1 ≠ 0$ .

Solutions: $x = 24$ or $10$ .

Case 3 : $A = -1$ , $B$ is even (so we have $(-1)^{2k} = 1$ )

$1 = A = x^2 - 17x + 71$

$\Rightarrow (x - 9)(x - 8) = 0$

If $x = 9$ , $B$ is the sum of one odd term and two even terms, so $B$ is odd.

If $x = 8$ , $B$ is the sum of three even terms, so $B$ is even.

Solutions: $x = 8$

There are no more cases; if $|A| ≠ 1$ and $B ≠ 0$ , simply consider the prime factorization of $A$ (as it is an integer) and no matter what B is (as long as not 0), you can't eliminate factors not equal to 1.

Our solutions are $x = 7, 10, 24,$ or $8$ , which sum to $49$ .

For $(x^2-17x+71)^{(x^2-34x+240)}=1$ , then either the following should hold: (1) base is equal to 1, (2) exponent is zero but the base is not equal to zero, (3) base is equal to $-1$ but the exponent should be even.

Case 1: base is equal to one

$\implies x^2-17x+71=1$

$\implies x^2-17x+70=0$

$\implies (x-10)(x-7)=0$

$\implies x=10$ or $x=7$

Case 2: exponent is zero

$\implies x^2-34x+240=0$

$\implies (x-10)(x-24)=0$

$\implies x=10$ or $x=24$

If $x=10$ or $x=24$ , the base will never be zero, hence $10$ and $24$ are solutions.

Case 3: base is equal to $-1$ but the exponent should be even

$x^2-17x+71=-1$

$\implies x^2-17x+72=0$

$\implies (x-8)(x-9)=0$

$\implies x=8$ or $x=9$

if $x=8$ , the exponent will be $8^2-34*8+240$ which is obviously even

if $x=9$ , the exponent will be $9^2-34*9+240$ which is obviously odd

hence only $x=8$ is a solution for this case

Thus, the solutions are $7, 10, 24, 8$ giving us a sum $7+10+24+8=49$ .

Because $x$ is an integer, the values of $a=x^2-17x+71$ and $b=x^2-34x+240$ must also be integers. Then the above relation becomes $a^b=1$ . Since $a$ and $b$ are integers, there are strict requirements on what they must be to satisfy this relation–

Case 1: $a=1$ , with no restriction on $b$ . Then we have $x^2-17x+71=1 \Rightarrow x^2-17x+70=0$ . Factorization gives us $x^2-17x+70=(x-7)(x-10)=0$ , so $x=7$ or $x=10$ .

Case 2: $a=-1$ , and $b$ is even. This gives us $x^2-17x+71=-1 \Rightarrow x^2-17x+72=0$ , and more factorization yields $x^2-17x+72=(x-8)(x-9)=0$ . The only possibilities for $x$ are $8$ and $9$ , but the latter choice makes $b=x^2-34x+240=(x-10)(x-24)=15$ , which is not even. $x=8$ makes $b=32$ even, so it's the only working choice in this case.

Case 3: $a\neq0$ , and $b=0$ . Because $71$ is prime, we can easily deduce that $a=x^2-17x+71=0$ has no integer solutions by the Rational Root theorem–we need only check $\pm71$ and $1$ , since they are the only factors of $71$ , and those clearly aren't solutions. So $a\neq0$ . As done in Case 2, factorizing $b$ gives $b=x^2-34x+240=(x-10)(x-24)=0$ , so $x=10$ and $x=24$ are solutions.

Altogether, $x=7$ , $x=10$ , $x=8$ , and $x=24$ are the only integers which satisfy $(x^2-17x+71)^{(x^2-34x+240)}=1$ . Summing these gives $7+10+8+24=49$ as the answer.

Exponent $= 0$ yields $x=10$ or $x=24$

$x^{2}-17x+71=1$ yields $x=7$ or $x=10$

$x^2-17x+71=-1$ yields $x=8$ or $x=9$ , of which only $x=8$ is permitted, because of an even exponent so 1 $0+24+7+8=49$

In order for the equation to be equal to 1, there are a few cases that can occur.

Case 1: Any integer to the power of 0 is equal to 1. Thus we put $(x^2 - 34x + 240) = 0$ . When we factor, we get $(x - 24)(x - 10) = 0$ , so $x = 24$ or $x = 10$ . We have to be careful because $0^0$ is undefined so we plug in the values of $x$ into $(x^2 - 17x + 71)$ and see if that equals 0. Since it does not, 10 and 24 are valid solutions.

Case 2: 1 to the power of any integer is equal to 0. So to get this case, we put $(x^2 - 17x + 71) = 1$ . Once we factor this equation, we get $(x - 10)(x - 7) = 0$ , so $x = 10$ or $x = 7$ .

Case 3 : Our last case is $(-1)^ N = 1$ , where $N$ is an even integer. We put our base, $(x^2 -17x + 71) = -1$ . Factoring this equation, we get $(x - 8)(x - 9) = 0$ so $x = 8$ or $x = 9$ . We need to verify that when we plug these numbers into $(x^2 - 34x + 240)$ we get an even number. Plug in 8, and the expression comes out to be 32, and even number. Plug in 9, and the expression comes out to be 15, an odd number. Thus 8 is the only valid solution in this case.

The problem asks for the sum of all the integer values of $x$ . So we add 10 + 24 + 7 + 8 which is equal to 49.

Let (x^2 - 17x + 71) = a, (x^2 - 34x +240) = b First case: if a = 1 , regardless of whatever b, a^b = 1. Thus, by solving a = 1, we get (x^2 - 17x + 71) = 1 \Rightarrow (x - 7)(x - 10) = 0. Thus, the solutions for x in this case are 7 and 10

Second case: if a = -1 , if b is even, a^b can be expressed as (a^2)^k = 1, which satisfies the condition. However, if b is odd, a^b can be expressed as (a^2)^k \times a = -1 , which does not satisfy the solution. Solving for (x^2 - 17x + 71) = -1, we get x= 8 or 9. By substituting 8 in b, we find that b is even, so 8 satisfies the condition. However, when 9 is substituted, b is odd, which does not satisfy the case. So, for this case, the only solution is 8.

Third case: a \neq 1 and a \neq -1 . For this case, b has to be 0. Thus, by solving for (x^2 - 34x + 240) = 0 , we get x = 10 or 24 as solutions. Note that 10 satisfies both the first and third case but should not be counted twice.

Sum of all integer values = 7 + 10 + 8 + 24 = 49

$x^a=1$ only if $x=1$ , or $x=-1$ when a is odd or $a=0$ when $x \neq 0$ . $x^2-34x+240=(x-24)(x-10)$ when $x^2-34x+240=0$ then $x=24,10$ . For these values , $x^2-17x+71\neq 0$ again, $x^2-17x+71={(x-7)(x-10)+1}$ if $x^2-17x+71=1$ then $(x-7)(x-10)=0$ . then $x= 7,10$ . if $x^2-17x+71=-1$ then $(x-7)(x-10)= -2$ . for integer $x$ there is only one solution and that is $x=8$ .for $x=8$ , $x^2-34x+240$ is even. so. the sum is $=24+10+7+8=49$

There can be three cases where the expression would yield a value of 1.

a. The exponent is zero, provided that the base is not zero. So, x^2 -34x + 240 = (x - 24)(x - 10) = 0 Hence, x = 24 or 10. Checking if both values would yield a base of zero, we have,

For x = 24, the base is equal to 24^2 - 17(24) + 71 = 239 which is not equal to zero. For x = 10, the base is equal to 10^2 - 17(10) + 71 = 1 which is not equal to zero. Hence, 24 and 10 are solutions.

b. The base is 1. So, x^2 - 17x + 71 = 1 x^2 - 17x +70 = (x - 7)(x - 10) = 0 Hence x = 7 or 10. Both of them are solutions.

c. The base is -1, provided that the exponent is even. So, x^2 - 17x + 71 = -1 x^2 - 17x + 72 = (x - 8)(x - 9) = 0 Hence, x = 8 or 9. Checking if the exponent is even, we have,

For x = 8, the exponent is 8^2 - 34(8) + 240 = 32 which is obviously even. For x = 9, the exponent is 9^2 - 34(9) + 240 = 15 which is obviously odd. Hence, 8 is also a solution.

Therefore, the sum of all the solutions is 24 + 10 + 7 + 8 = 49. QED

For the above question, we know result can be obtained only in four ways so we have to consider 4 cases

Case 1 : when $x^2 - 34x +240 = 0$ and on solving the quadratic equation we get the possible solutions of x are $x= 10 or 24$ . These are 2 of the solution

Case2: when $x^2 -17x + 71=0$

(here one fact to notice is for an integer value of x both $x^2 - 34x +240 = 0$ and $x^2 -17x + 71=0$ and will not exist simultaneously)

Here we see that integer roots of the equation does not exist .So this case is not valid

Case 3: When $x^2 -17x + 71= 1$ ,
$x^2 - 17x + 72= 0$

On solving we get the two possible solutions are $7 or 10$

Case 4: When $x^2 -17x + 71= -1$ , and here $x^2 - 34x +240$ should be even .

On solving $x^2 -17x + 71= -1$ we get the 2 solutions are $8 and 9$ and on substituting this value in $x^2 - 34x +240$ we get an even value only in case of 8 .

So in this case the possible solution is 8, Adding up,

$24 + 10 + 7 + 8 =49$

We have: $x^2 - 34x + 240 = (x-10)(x-24),$ $x^2 - 17x + 71 = (x-7)(x-10)+1 = (x-8)(x-9)-1$

Since $a^0=1 \, \forall a \neq 0, \quad 1^b=1 \, \forall b$ and $(-1)^c=1 \, \forall c$ even, we find $x=7,8,10,24$ satisfy our equation. Thus $7+10+24+8=49$ .

The only possible solutions occur when $x^2-34x+240=0$ or $x^2-17x+71=\pm1$

However, not all of the above cases are solutions, so we will need to check the cases.

The solutions to $x^2-34x+240=0$ are $x=10, 24$ . For both these cases, $x^2-17x+71\neq0$ so both are valid solutions.

The solutions to $x^2-17x+71=1$ are $x=7,10$

The solutions to $x^2-17x+71=-1$ are $x=8, 9$ . When $x=8$ , $x^2-34x+240=32$ and the resulting value is $1$ . When $x=9$ , $x^2-34x+240=15$ and the resulting value is $-1$ . Therefore $x=9$ is not a valid solution.

The valid solutions are $x=7, 8, 10, 24$ and thus the sum is $49$

the given equation is $(x^{2}- 17x +71)^{(x^{2}-34x+240)}= 1$ implies that value of equations should be of form...

1. $1^{k} => x^{2}- 17x +71 = 1$ k is real no.solving we get x=10,x=7

2. $(-1)^{k}=> x^{2}- 17x +71 = -1$ and $x^{2}-34x+240=k$ (k is even) solving $x^{2}- 17x +71 = -1$ we get x=8,x=9 but for 9 $x^{2}-34x+240$ is odd. So ,x=8 is a solution.

3. $k^{0}=> x^{2}- 17x +71 \neq 0$ and $x^{2}-34x+240=0$ solving $x^{2}-34x+240=0$ We get x=24 and x=10 and $x^{2}- 17x +71 \neq 0$ for 24 or 10 .

So, total solutions are x= 10,8,7,24 sum=49

any number raised to power zero is 1 excepting zero itself. equating power to 0 we get x = 10,24. for these the base is non zero.

1 raise to the power of any number is 1 .. we get equating base=1, x = 7,10

-1 raised to the power of even numbers give 1; -1 = base gives x =8,9 but only x = 8 yields even power...

solution: x = 7,8,10,24

let (x^2-17x+71)=1 and also let it = -1 x^2-34x+240= 0

let x^2-34x+240 be y

(x^2-17x+71)^y*1/y = 1^1/y

x^2-17x+71= +1 or -1

IF WE SOLVE THIS WE WILL GET 4 VALUES FOR X. SIMILARLY IF WE EQUATE THE POWERED POLYNOMIAL TO 0 WE WILL GET 2 VALUES FOR X.

THE VALUES OF X WILL BE 7,8,9,10,24. IN THIS SET 9 WILL NOT BE APPLICABLE AS IT DOESA NOT HOLD GOOD FOR ALL EQUATIONS.

SUM OF ALL VALUES OF X=7+8+10+24=49

We consider the following cases:

Case 1: Base is 1, exponent is any value. In this case, $x^2-17x+71 = 1\Rightarrow (x-7)(x-10)=0$ has solutions $x=7, 10$ .

Case 2: Base is -1, and exponent is even. In this case, $x^2 - 17x + 71 = -1\Rightarrow (x-9)(x-8) = 0$ . The exponent is even if and only if $x$ is even, so $x=8$ is the only solution.

Case 3: Base is not 0, exponent is 0. In this case, $x^2-34x+240 = 0 \Rightarrow (x-10)(x-24)=0$ . A quick check shows that the base is not 0 for both values, so $x = 10, 24$ are solutions.

(Note that $x=10$ appears twice as a solution, so we may not simply use Vieta's formulae in Case 3.) Thus the sum of all real values of $x$ is $7+8 +10 + 24 = 49$ .

The scenarios in which the result of the LHS will be 1 are:

• $1^a$ , where a is any value
• $(-1)^e$ , where e is an even value
• $0^n$ , where n is any non zero value

Now, the roots of $x^2 - 34x + 240 = 0$ are $10$ and $24$ , which are obviously not the roots of $x^2 - 17x + 71 = 0$ . So, we don't have the problem of $0^0$ here. Thus, $x=10, 24$ are two definite solutions.

If, $x^2 - 17x + 71 = 1$ , the solutions are $x = 7, 10$ .

If, $x^2 - 17x + 71 = -1$ , the solutions are $x = 8, 9$ . But the power has to be even. Now, if we look at the power, $x^2 - 34x + 240$ , the parts $34x$ and $240$ are even as $x$ is an integer. So, $x^2$ must be even or rather $x$ must be even . Thus only $x = 8$ works.

So the distinct solutions are: $x = 7, 8, 10, 24$

Sum of the possible values is: $49$

Keep alert about 10....here 10 is come two times so count it only one time..

Let $f\left(x\right) = x^2 - 17x + 71$ and let $g\left(x\right) = x^2 - 34x - 240$

We require ${f\left(x\right)} ^ { g\left(x\right)} = 1$ , and so we have three cases:

*Case 1 : * $f\left(x\right) = 1, \;g\left(x\right)$ is arbitrary

*Case 2 : * $f\left(x\right) = -1, \; g\left(x\right)$ is even

*Case 3 : * $f\left(x\right) \neq 0, \; g\left(x\right) = 0$

Now for the first case:

$f\left(x\right) = 1 \Rightarrow x^2 - 17x + 70 = 0$

$\Rightarrow \left(x-7\right)\left(x-10\right) = 0 \Rightarrow x = 7,$ or $x=10$ . Recall that the value of $g\left(x\right)$ here is arbitrary.

For the second case:

$f\left(x\right) = -1 \Rightarrow x^2 - 17x + 72 = 0$

$\Rightarrow \left(x-8\right)\left(x-9\right) = 0 \Rightarrow x = 8,$ or $x= 9$

Note however $34$ and $240$ are both even and thus $g\left(x\right)$ is even $\Leftrightarrow x$ is even.

So we need only consider $x = 8$

Finally, for the third case:

$x^2 - 34x + 240 = 0 \Rightarrow \left(x - 10\right)\left(x-24\right) = 0$

$\Rightarrow x = 10$ or $x = 24$

Note that since $71$ is prime, $f\left(x\right) = 0$ cannot have integer solutions. Thus we have that roots:

$x \in \{7,8,10,24\}$

This gives the required answer of $7 + 8 + 10 + 24 = 49$

It is 49 because when you solve it is 1 or the exponent is zeeerooo or itis miinus one and then the exponent has be 2 or another even thing

This equation can be simplified to

$((x-8)(x-9)-1)^{(x-24)(x-10)} = 1$ [1]

or

$((x-7)(x-10)+1)^{(x-24)(x-10)} = 1$ [2]

Obviously $x = 24, 10$ are solutions, and the base doesn't equal $0$ .

Consider $1^{a} = 1$ for all $a$ .

From [2] we get $(x-7)(x-10) = 0$ , which gives $x = 7, 10$ . (no need to check [1] because 2 equations are equivalent.]

Consider $(-1)^{a} = 1$ for all even integers $a$ .

From [1] we get $(x-8)(x-9) = 0$ , which gives $x = 8, 9$ , but $x = 9$ doesn't make $(x-24)(x-10)$ even integer. Therefore, $x = 8$ .

Therefore, the solutions are $\boxed{x = 7,8,10,24}$ .

We solve the equation ln(x^2-17x+71) (x^2-34x+240)=ln1=0 we find the roots: 7, 10 and 24. Equations ln(x^2-17x+71) =0 and x^2-34x+240=0 have the common root 10. Since 0^0 is undefined we discard this solution. Note that (-1)^(even integer)=1, so we solve the equations x^2-17x+71=-1 and we find the roots: 8 and 9. For x=9 the expression x^2-34x+240 takes the value 15, so we discard 9 as a solution since 15 is an odd integer. For x=8 the expression x^2-34x+240 takes the value 32 which is an even integer. So we accept 8 as a solution. Adding all the previous solutions we get the desired sum.