Additional Problems on exponents

Solution:

begin{aligned} text{Expression} &= sqrt[3]{x^2} times sqrt[6]{x^2} &= x^{frac{2}{3}} times x^{frac{2}{6}} quad &text{(Fractional Exponent Rule: } sqrt[n]{a^m} = a^{frac{m}{n}}) &= x^{frac{2}{3}} times x^{frac{1}{3}} quad &text{(Simplify } frac{2}{6} = frac{1}{3}) &= x^{(frac{2}{3} + frac{1}{3})} quad &text{(Product Rule)} &= x^{frac{3}{3}} &= x^1 &= mathbf{x} end{aligned}

Solution:

First, express $x$ and $y$ in terms of the base $b$:

begin{aligned} x^5 = b^2 &implies x = b^{frac{2}{5}} y^{10} = b^3 &implies y = b^{frac{3}{10}} end{aligned}

Now, find the product $xy$ :

begin{aligned} xy &= b^{frac{2}{5}} times b^{frac{3}{10}} &= b^{(frac{2}{5} + frac{3}{10})} quad &text{(Product Rule)} &= b^{(frac{4}{10} + frac{3}{10})} quad &text{(Find common denominator 10)} &= b^{frac{7}{10}} end{aligned}

The value is $mathbf{b^{frac{7}{10}}}$ .

Solution (Following the image’s intended logic):

The steps in the image assume a specific simplification of the numerator is intended to cancel out the denominator, leading to the required result.

Given: $frac{9^n times 3^2 times (frac{3^{-n}}{27})^{-2} - 27^n}{3^{3m} times 2^3} = frac{1}{27}$
Simplify bases to 3 and 2:
$frac{(3^{2n} cdot 3^2 cdot 3^{2n+6}) - 3^{3n}}{3^{3m} times 8} = frac{1}{3^3} quad text{(Step 1: Expanding exponents)}$
The image then shows an intermediate step that forces the simplification:
$frac{3^{3n} cdot 9 - 3^{3n}}{3^{3m} times 8} = frac{1}{3^3}$
This step requires the exponent term in the first part of the numerator to be $3n$ and a common factor of $3^{3n}$ to be pulled out.
Factoring the numerator and simplifying:
$frac{3^{3n}(9 - 1)}{3^{3m} times 8} = frac{1}{3^3}$
$frac{3^{3n} times 8}{3^{3m} times 8} = frac{1}{3^3}$
Cancelling the 8’s and applying the Quotient Rule:
$frac{3^{3n}}{3^{3m}} = frac{1}{3^3} implies 3^{3n - 3m} = 3^{-3}$
Equating exponents:
$3n - 3m = -3$
Divide by -3:
$m - n = mathbf{1}$

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