Question

Someone please give a step by step way on how to solve this problem

Answer 1

10

Explanation:

There are several ways to solve this problem which I’ll demonstrate these ways and see which method is way easier to understand.

First Method

This method is to convert the fractional expression to a single exponent expression using a law of exponent.

`\displaystyle \large{(m^a)/(m^b) = m^(a-b)}`

In words, it means if both divisor (denominator) and dividend (numerator) have same values or are same, we can subtract the exponent from dividend (numerator) with divisor (denominator).

The dividend (numerator) exponent is unknown, let’s say it’s got ‘x’ as an exponent and this ‘x’ is what we are going to find since we don’t know its value.

Now we have the equation:

`\displaystyle \large{(m^x)/(m^7) = m^3}`

Apply the exponent rules above.

`\displaystyle \large{m^(x-7) = m^3}`

Solve an exponential equation - If both sides of equation have same base, we can solve the exponents.

Since LHS (Left-Handed Side) and RHS (Right-Handed Side) both have same base which is ‘m’ therefore, solve exponents.

`\displaystyle \large{x-7=3}\\\displaystyle \large{x=7+3}\\\displaystyle \large{x=10}`

Therefore, the missing exponent is 10.

Additional Info

You can also use logarithm to solve an exponential equation.

`\displaystyle \large{m^(x-7) = m^3}`

Take a logarithm of base ‘m’ both sides so we can apply other properties.

`\displaystyle \large{\log_m m^(x-7) = \log_m m^3}`

From the property
`\displaystyle \large{\log_a b^n = n \log_a b}`

`\displaystyle \large{(x-7) \log_m m = 3 \log_m m}`

From the property
`\displaystyle \large{\log_a a = 1}`

`\displaystyle \large{(x-7)1 = 3 \cdot 1}\\\displaystyle \large{x-7=3}\\\displaystyle \large{x=10}`

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Second Method

This method is to isolate the term with missing exponent.

`\displaystyle \large{(m^x)/(m^7) = m^3}`

Transport m^7 to multiply with m^3 via transportation property.

`\displaystyle \large{m^x = m^3 \cdot m^7}`

We also have to apply another law of exponent as well.

`\displaystyle \large{m^a \cdot m^b = m^(a+b)}`

Similar to the one with division (fraction), this one, when same bases multiply each other, we have to add both exponents while the other one subtracts exponents.

`\displaystyle \large{m^x = m^(3+7)}\\\displaystyle \large{m^x = m^(10)}`

In the first method, we know that when solving an exponential equation, if two bases from left and right sides are same, we solve the exponent.

Therefore,
`\displaystyle \large{m^x = m^(10) \to \boxed{x=10}}`

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Summary

Laws of Exponent

`\displaystyle \large{a^m \cdot a^n = a^(m+n)}\\\displaystyle \large{(a^m)/(a^n) = a^m / a^n = a^(m-n)}\\`

For what I demonstrate in my explanation only.

Exponential Equation

When two bases from left side and right side are same, we can solve through exponents.

Examples:

`\displaystyle \large{2^(x-3) = 2^(4)}\\\displaystyle \large{x-3=4}\\\displaystyle \large{x=4+3}\\\displaystyle \large{x=7}`

You can also substitute x-value in an equation to see if both sides are equal or not.

`\displaystyle \large{2^(x-4) =2^3}`

Substitute x = 7 for check.

`\displaystyle \large{2^(7-4) = 2^3}\\\displaystyle \large{2^3 = 2^3 \ \ \checkmark}`

Logarithm

This is not necessary used in this question. This is mostly used for converting to exponential and more advanced mathematics.

`\displaystyle \large{\log_a b^n = n\log_a b}\\\displaystyle \large{\log_a a = 1}`

Only from what I demonstrate in my explanation.

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If you are still stuck or need clarification regarding my answer and explanation, you may ask in comment below.

Hope this helps! And wish you luck in your assignment.