Wrote a division problem which two numbers written in scientific notation have a quotient of 5•10^-5

Answer 1

Let's assume we have two numbers written in scientific notation:

Number 1: a₁ x 10^b₁
Number 2: a₂ x 10^b₂

The quotient of these two numbers is:

(a₁ x 10^b₁) ÷ (a₂ x 10^b₂)

To obtain a quotient of 5 x 10^-5, we need to set up the equation:

(a₁ x 10^b₁) ÷ (a₂ x 10^b₂) = 5 x 10^-5

Now, let's find the values of a₁, a₂, b₁, and b₂:

Since the base of the exponents is the same (10), we can combine the exponents:

(a₁ ÷ a₂) x 10^(b₁ - b₂) = 5 x 10^-5

To simplify the calculation, we can divide both sides by 5:

(a₁ ÷ a₂) x 10^(b₁ - b₂) ÷ 5 = 10^-5

Now, we can distribute 10^-5 to the numerator and simplify the equation further:

(a₁ ÷ a₂) x (10^(b₁ - b₂) ÷ 5) = 1 x 10^0

At this point, we want to simplify the equation more by setting the exponent of 10 to zero. So:

10^(b₁ - b₂) ÷ 5 = 1

To make it clearer, let's rewrite the equation:

10^(b₁ - b₂) = 5

Now, we need to find the values for b₁ and b₂ that satisfy this equation. We can start by assuming a value for b₂, and then find the corresponding value for b₁:

Let's assume b₂ = -5:

10^(b₁ - (-5)) = 5
10^(b₁ + 5) = 5

To solve for b₁, we can take the logarithm (base 10) of both sides:

log₁₀(10^(b₁ + 5)) = log₁₀(5)
(b₁ + 5) = log₁₀(5)

To isolate b₁, we subtract 5 from both sides:

b₁ = log₁₀(5) - 5

So, a possible division problem with two numbers in scientific notation that has a quotient of 5 x 10^-5 could be:

(a₁ x 10^(log₁₀(5) - 5)) ÷ (a₂ x 10^-5) = 5 x 10^-5