Is there an efficient way of testing if the resulting value of an exponential gives an integer without actually expanding the equation. For example: log(12)−log(4)=1.09861…and is a transcendental number. But if we take the exponential, we get an integer: e^(log(12)-log(4))=e^(1.09861...)=3 This can be easily shown by proving that the equation is equivalent to 12/3. However is there a way to prove this while staying in the logarithm nation... We are assuming that the two numbers 12 and 4 are very large...Can I find say numerical stability bounds to this simple equation: 2.71828^(2.48491-1.38629)=3...such that if I use n digits, I will get an integer close the the n-th digit. This may look trivial, but let's assume we do not have say log(12), but only 2.48491… up to some precision of k-digit

$\frac{20b}{{\left(4b^3\right)}^3}$

Which operation could we perform in order to find the number of milliseconds in a year??
${\displaystyle 60\cdot 60\cdot 24\cdot 7\cdot 365}$
${\displaystyle 1000\cdot 60\cdot 60\cdot 24\cdot 365}$
${\displaystyle 24\cdot 60\cdot 100\cdot 7\cdot 52}$
${\displaystyle 1000\cdot 60\cdot 24\cdot 7\cdot 52?}$

Tell about the meaning of Sxx and Sxy in simple linear regression,, especially the meaning of those formulas

Is the number 7356 divisible by 12? Also find the remainder.
A) No
B) 0
C) Yes
D) 6

What is a positive integer?

Determine the value of k if the remainder is 3 given ${\displaystyle (x^3+k{x}^2+x+5)÷(x+2)}$

Is $41$ a prime number?

What is the square root of $98$?

Is the sum of two prime numbers is always even?

149600000000 is equal to
A)$1.496\times <!--\; \times \; -->{10}^{11}$
B)$1.496\times <!--\; \times \; -->{10}^{10}$
C)$1.496\times <!--\; \times \; -->{10}^{12}$
D)$1.496\times <!--\; \times \; -->{10}^8$

Find the value of$\log 1$ to the base $3$ ?

What is the square root of 3 divided by 2 .

write ${\displaystyle \sqrt[5]{{\left(7x\right)}^4}}$ as an equivalent expression using a fractional exponent.

simplify ${\displaystyle \sqrt{125n}}$

What is the square root of ${\displaystyle \frac{144}{169}}$