Solving Equations on exponents

Solution:

begin{aligned} text{Given: } 4^x &= (2sqrt{2})^8 (2^2)^x &= (2^1 cdot 2^{frac{1}{2}})^8 quad &text{(Express all bases as 2)} 2^{2x} &= (2^{frac{3}{2}})^8 quad &text{(Product Rule: } 1 + frac{1}{2} = frac{3}{2}) 2^{2x} &= 2^{(frac{3}{2} times 8)} quad &text{(Power of a Power Rule)} 2^{2x} &= 2^{12} 2x &= 12 quad &text{(Equating exponents)} x &= 6 x+2 &= 6+2 = mathbf{8} end{aligned}

Solution:

begin{aligned} text{Given: } frac{2^{5x}}{2^x} &= sqrt[5]{2^{20}} 2^{5x-x} &= 2^{frac{20}{5}} quad &text{(Quotient Rule and Fractional Exponent Rule)} 2^{4x} &= 2^4 4x &= 4 quad &text{(Equating exponents)} x &= mathbf{1} end{aligned}

Solution:

begin{aligned} text{Given: } 5^{x-2} times 3^{2x-3} &= 135 5^{x-2} times 3^{2x-3} &= 5^1 times 3^3 quad &text{(Express 135 as } 5 times 27 = 5^1 times 3^3) end{aligned}

By comparing the exponents of the corresponding bases:

For base 5:

x-2 = 1 implies x = mathbf{3}

For base 3:

2x-3 = 3 implies 2x = 6 implies x = mathbf{3}

The value is $\mathbf{x=3}$.

Solution:

begin{aligned} text{Given: } 3^x &= 270 3^{x-3} &= frac{3^x}{3^3} quad &text{(Quotient Rule in reverse: } a^{m-n} = frac{a^m}{a^n}) 3^{x-3} &= frac{270}{27} quad &text{(Substitute given value and } 3^3=27) 3^{x-3} &= mathbf{10} end{aligned}

Solution:

Let $a^x = b^y = c^z = K$ (where $K$ is a constant).

Using the rule $a^m=b^n \Rightarrow a=b^{n/m}$:

a = K^{frac{1}{x}}, quad b = K^{frac{1}{y}}, quad c = K^{frac{1}{z}}

Now, substitute these expressions into the second given equation, $b^2 = ac$ :

begin{aligned} (K^{frac{1}{y}})^2 &= (K^{frac{1}{x}}) times (K^{frac{1}{z}}) K^{frac{2}{y}} &= K^{(frac{1}{x} + frac{1}{z})} quad &text{(Power of a Power Rule and Product Rule)} end{aligned}

Equating the exponents since the bases are the same:

mathbf{frac{2}{y} = frac{1}{x} + frac{1}{z}}

The statement is proven.

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