Powers between near powers.

To find the value of x, we will have to take 1-7th power of all terms in the inequality. But it is not easy to find the value of $(5^{21}+1)^{1/7}$ or $(2^{49}-1)^{1/7}$ . But the main thing to notice in this problem is that we can easily find the seventh roots of $5^{21}$ and $2^{49}$ . We must somehow use that in the problem.

Take each inequality individually:

1. $x^{7}<2^{49}-1$ Since $x^{7}$ is less than $2^{49}-1$ , It must also be less than $2^{49}$ which is greater than $2^{49}-1$ .

$x^{7}<2^{49}$ $x<2^{7} or x<128$

But x is actually less than $(2^{49}-1)^{1/7}$ which should be around 127.9 (Mathematical Intuition). Still, the integral values of x are less than 128.

2. $x^{7}>5^{21}+1$ Since $x^{7}$ is greater than $5^{21}+1$ it must also be greater than $5^{21}$ which is less than $5^{21}+1$ .

So, $x^{7}>5^{21} or x>5^{3} or x>125$ But x is actually greater than $(5^{21}+1)^{1/7}$ which should be around 125.1 (Mathematical Intuition). Still, the integral values of x are greater than 125.

So let's combine: $125<x<128$ So x can be 126 or 127. So answer is 2

5^21 = 5^(3.7)=125^7; 2^49 = 2^(7.7) = 128^7; 125^7 + 1 < X^7 < 128^7 - 1<=> 125 < X < 128 <=> X = 126, 127. There is 2

If we think that:

$x^{7} > 5^{21} + 1 = x > 5^{\frac{21}{7}} + 1^{\frac{1}{7}}$

> By definition: $1^{n} = 1$

$x > 5^{3} + 1$

$x > 125 + 1 = \boxed{x > 126}$

And:

$x^{7} < 2^{49} - 1 = x < 2^{\frac{49}{7}} - 1^{\frac{1}{7}}$

$x < 2^{7} - 1$

$x < 128 - 1 = \boxed{x < 127}$

Well... we know that 126 < x < 127 !

" x " is between these two integers... then there will be 2 integers satisfying " x "

• 1 -> 126

• 2 -> 127

  $= \boxed{2}$

All we have to do is to evaluate the inequalities separately:

First inequality , $5^{21}+1< x^{7}$ .We may rearrange the term and see that $(5^{3})^{7}-x^{7}< -1$ . It is obvious that the integer values satisfying the inequality is $x=126,127,...$

Second inequality , $x^{7}< 2^{49}-1$ . Same as the former, we arrange the terms and see that $1< (2^{7})^{7}-x^{7}$ . It is also obvious that the integer values satisfying this inequality is $x=127,126,...$ .

Hence, the value of $x$ satisfying both inequalities is only $126$ and $127$ . There are $\boxed{2}$ solutions.

Hello,

as 5^(21) +1 < x^7 < 2^(49) - 1

x^7 > 5^(21) + 1(1st condition)

x^7 = (125)^7

x = 125,

x^7 = 2^(49) -1

x^7 = (128)^7(2nd condition)

x=128

as the range 125 < x < 128 = (125)^7 < x^7 < (128)^7 = 5^(21) + 1 < x^7 < 2^49 - 1,it has only have 2 solutions as x = 126,127....therefore,total number of solutions is 2....

using the calculator:

5^{21} +1 < x^{7} < 2^{49} -1 ( take the seventh root from both sides ) "by the calculator as well"

125 < x < 128

then the possible integer solutions are 126 ans 127

which are 2 solutions