Seven Fifty Seven

$7^{x+1}+7^{x-1}=50\\ 7^x\times7+\frac{7^x}{7}=50 \\7^x(7+\frac{1}{7})=50 \\7^x(\frac{50}{7})=50 \\7^x=7 \\x=\log_7 7 \\x=\boxed{1}$

$7^{x+1}+7^{x-1} = 50$

$7^{x+1}+7^{x-1} = 7^{2}+7^{0}$

$x+1+x-1 = 2$

$2x = 2$

$x = 1$

$7^{x+1}+7^{x-1}=7^{x-1}(7^2+1)=50*7^{x-1}=50$ $\Rightarrow x-1=0, x=\boxed{1}$

Here is an arguably more intuitive and efficient way to solve this problem at a glance.

You know that the difference between $7^{x+1}$ and $7^{x-1}$ is $7^{2}$ because:

$(x+1) - (x-1) = 2$ thus,

$7^{2} = 49$

So the only way for $7^{x+1} + 7^{x-1} = 50$ is if

$7^{x-1} = 1$

Therefore:

$x-1 = 0$

$x=1$

7^x+1^ + 7^x-1^ = 50

7^x^ * 7 + 7^x^/7 = 50

7^x ^(7 + 1/7) = 50

7^x^ * 50/7 = 50

7^x^ = 7^1^

X = 1

It's really simple. If you know that 7 to the power of 2 is 49, then obviously you have to add 1 more to 49. So if you plug in 1 for x, then you get 7 to the power of 2 (49) + 7 to the power of 0 (1) which equals 50!

7^(x +1) + 7^(x-1) =50

or, 7^x . 7 + 7^x . 1/7=50

or, 7^x( 7 + 1/7 )=50

or, 7^x . 50/7 =50

or, 7^x = 7 ^1

or, x =1

You could use logs but honestly I just guessed and checked. Worked great.

Why do people not check their work in case of any extraneous solutions were found? Also, how many would have solved it like how I did?

how about this one?:)

7^x+1 + 7^x-1 = 50

[(x+1) + (x-1)]log(7) = log(50)

2x = log(50)/log(7)

x = 1

7^(x+1) + 7^(x-1) = 50

7^(x-1) (7^2 + 1) = 50

7^(x-1) = 1

x - 1 = 0

x = 0

Another solution

7^(x+1) + 7^(x-1) = 50

7^(x+1) + 7^(x-1) = 49 + 1

7^(x+1) + 7^(x-1) = 7^2 + 7^0

x+1 = 2 and x-1 = 0

Then

x = 1

A very simple problem. 7 ^(x+1)+ 7^(x-1). x=1 7^(1+1) + 7^ (1-1)= 49+1=50. I am surprised that no other kids can solve this, only adults.

$7^{x+1}+7^{x-1}=50 \implies 7^2 \times 7^{x-1}+7^{x+1} \implies (49+1)7^{x-1}=50 \implies 50 \times 7^{x-1}=50 \implies 7^{x-1}=1$

$\implies 7^{x-1}=7^0 \implies x-1=0 \implies x=\boxed{\large{1}}$

• $7{x + 1} + 7{x - 1} = 50$
• $7{x} \cdot 7 + 7{x} \cdot 7{-1} = 50$
• $7{x}(7 + 7^{-1} = 50$
• $7^{x} \cdot \frac{50}{7}$ $= 50$
• $7^{x} \cdot \frac{1}{7}$ = $1$
• $7^{x - 1}$ = $7^{0}$ $x - 1 = 0, x = 1$

Get coefficients

No one likes a fraction there

Divide 50 and solve power

7^x+1 + 7^x-1 = 50 7^x+1 + 7^x-1 = 7^2 + 7^0 (x+1)+(x-1)=2 + 0 2x = 2 x=1

7^0=1 ,7^1=7, 7^2=49 then: 1 + 49=50

Divid expression by 7 so 7^(x +1)\7+7^(x -1)\7=50\7,,,7^(x+1-1)+7^(x-2)=50\7,,,,,,,,, 7^x+7^(x)×7^-2=50\7,,7^x×(1+7^-2)=50\7,,7^x×(50\7^2)=50\7,,7^x=7^1,,,x=1###

7^(x-1)(7^2 +1) = 50 7^(x-1)(50) = 50 7^(x-1) = 1 x-1 = 0, because 7^0=1 Therefore x = 1

7^(x+1) + 7^(x-1) =50

Solving the exponent

(x+1)(x-1)

x^2 - 1^2

x=√1

x=1

7^(x+1)+7^(x-1)=50

7^(x+1)+7^(x-1)=7^2+7^0

Equating differently

7^(x+1)=7^2 or 7^(x-1)=7^0

=>(x+1)=2 or (x-1)=0

=>x=1

(7^x.7^1) +(7^x/7^1) = 50 |.7

(7^x.7^2) +(7^x) = 350

(7^x)(7^2 + 1) = 350

7^x (50) = 350

7^x = 350/50

7^x = 7^1

x = 1

$7^{x+1} + 7^{x-1} = 50$

$7^{x-1}(7^2 + 1) = 50$

$7^{x-1}(49 + 1) = 50$

$7^{x-1}\times50 = 50$

$7^{x-1} = 1$

$7^{x-1} = 7^0$

$x-1 = 0$

$\boxed{x = 1}$

(7^x+1)+(7^x-1)=50

OR(7^x)*7 + (7^x)/7=50

OR(7^x)(7+1/7)=50

OR(7^x)(50/7)=50

OR 7^x=50*7/50=7=7^1

Therefore, x=1

Call 7^{x+1} = y; And remember that 7^{x-1} = 7{(x+1) -1 -1} = 7^{x+1} \times 7{-2}; So, 7^{x+1} + 7^{x-1} = 50 it's the same thing of y + y \times 7{-2} = 50; y = 49; If 7^{x+1} = y; Then, 49 = 7^{x+1}; 49 = 7^{2} = 7^{x+1}; x+1 = 2; x = 1.

7^(x+1) + 7^(x-1) = 50 so

7^(x+1) + 7^(x-1) = 49 + 1

rewriting the expression with 7 as its base:

7^(x+1) + 7^(x-1) = 7^2 + 7^0

By properties of exponential function

(x+1)+(x-1) = 2+0

x+x+1-1=2

2x=2

x=1

7^(x+1) +7^(x-1)=7^2 +7^0

x+1+x-1 =2+0

2x =2

x=1

Do we have some other method to solve this problem

X+1=2 x-1=1. 7x7=49+1=50, so X=1

{ 7 }^{ x+1 }={ 7 }^{ x }X7\ 7^{ x-1 }=\frac { { 7 }^{ x } }{ 7 } \ \ { 7 }^{ x }(7+\frac { 1 }{ 7 } )=50\ { 7 }^{ x }=7\quad \quad (7+1/7=50/7)\ x=1

Guess and check

7^{x+1} + 7^{x-1} = 50

this can be written as

7^{x+1} + 7^{x-1} = 7^{2} + 7^{0}

7^{x+1} + 7^{x-1} = 7^{1+1} + 7^{1-1}

So, x = 1

7^(x+1)+7^(x-1)=50 or, 7^x(7+1/7)=50 or, 7^x*(50/7)=50 or, 7^x=7=7^1 or, x=1

Am i the only one one who just thought "hey 7^2=49 7^0 =1 49+1= 50 !" 7^2 = 7^ 1 +1 = 49 7^0 =7^ 1 -1 = 1

This method is more applicable to a wider range of problems:

$7^{2x}=50$ $2x \times ln{7} = ln{50}$ $2x = \frac{ln{50}}{ln{7}}$ $x= \boxed{1}$

Ok, I think I cheated, used logic rather than math. I know 7^2 is 49, very near to the solution. So I thought it might help if I write 50 as terms that involve 7 to a certain power. 50=49+1=7^2+7^0. Well if I look to the other side, it is extremely easy make the exponents equal 2 and 0. So x must be 1.

$7^{x+1}$ + $7^{x-1}$ = $7^{2}$ +1

$7^{x+1}$ + $7^{x-1}$ = $7^{2}$ + $7^{0}$

by analogy it's very clear that $\boxed{x=1}$

or

$7\times7^{x}+\frac{7^{x}}{7}=50$

$7^{x}\times(7+\frac17)=50$

$7^{x}\times(\frac{50}7)=50$

$50\times7^{x-1}=50$

$7^{x-1}=1$

$Log_{7}(7^{x-1})=Log_{7}(1)$

$x-1=0$

$\boxed{x=1}$

7^(x+1)+7^(x-1)=50

>7^x*7+7^x / 7=50 >let 7^x =a then the equation is simply

>7 a + a / 7 =50 >(49+1)a / 7 = 50 >50 a / 7 = 50 >a = 7

>putting the value of a = 7^x we find >7^x = 7 >x = 1 (using exponential rule : if a^x = a^b then x = b)

The t possibilities for getting 50 with powers of 7 are just with 7 power 0, 7 power 1, and 7 power 2. Logically 7 power 0 + 7 power 2 gives the result, as 49+1 = 50. now equate it as x+1=2 and x-1=0. You will get answer as 1.

I have directly put the value x=1 because it is simple. Only you have to find the relation in between 7 and 50 such as
50 = 49 +1
Since 49 = square of 7 then we have to put only 1 and we can get the answer.

7^x+1 +7^x-1=50 7^2x=50 49x=50 X=50-49 X=1

This is a case of converting the number bases.

Since the left hand side is only in the form of sum of powers of $7$ , this leads me to convert the number $50$ to base $7$ .

Listing down the first few powers that don't exceed $50$ , we have $7^0 = 1,7^1 = 7, 7^2= 49$ , in which case we can see that $49 + 1 = 50$ or $7^2 + 7^0 = 50$

So we compare: $x+1 = 2, x - 1 = 0$ which gives $x = \boxed{1}$

now we know we can take 7^2 + 7^0=49+1=50 so if we take x as 1 then 1+1=2 and 1-1=0 SOLVED......

7^(x+1)+7^(x-1)=50

=> 7^x.7+7^x/7=50

=> 7^x (7+1/7)=50

=> 7^x(50/7)=50

=> 7^(x-1)=1

=> x-1 = 0

=> x = 1

$7^{x+1}+7^{x-1}=7^{x}7^{1}+7^{x}7^{-1}=7^{x}(7+\frac{1}{7})=50$

Rewrite $7+\frac{1}{7}$ as $\frac{49}{7}+\frac{1}{7}=\frac{50}{7}$ :

Therefore $7^{x}(\frac{50}{7})=50$

$\Rightarrow 7^{x}=7 \Rightarrow x=1$

7^x-1(7^2 + 1) = 50 7^x-1(50) = 50 7^x-1(50)/50 = 50/50 7^x-1 = 1 7^x-1 = 7^0 x-1 = 0 x = 1

7^x(7+1/7)=50 7^x(50/7)=50 7^x=7 Xlog7=log7 X=1

(X+1)+ (x-1)= 7^2+1=7^2+7^0 2x=2+0=2. Hence x=1.

its logic.... (x+1)-(x-1)=2 7^2=49 x=2

7^x(7+1/7) =50

7^x(50/7)=50

7^x=7

Hence x=1

50 = 49 - 1, 50 = 7^2 + 7^1, if 7^(x+1)+7^(x-1)=50, then x+1=2 and x=1

Suppose 7^x=a then a/7 +7a=50 => a/7 + (49a)/7=50 => 50a=350 => a=7 => 7^x=7 => x=1