If 4^{a - 1} \cdot 8^{2a} = 16^3, what is the value of a? — SAT

If $4^{a - 1} \cdot 8^{2a} = 16^3$ , what is the value of $a$ ?

\\frac{7}{4}

The correct answer is

\frac{7}{4}

. Rewriting each base in the equation

4^{a - 1} \cdot 8^{2a} = 16^3

as a power of 2 gives

(2^2)^{a-1} \cdot (2^3)^{2a} = (2^4)^3

. Applying the power of a power rule results in

2^{2a-2} \cdot 2^{6a} = 2^{12}

. Using the product rule of exponents to combine the left side yields

2^{2a-2+6a} = 2^{8a-2} = 2^{12}

. Setting the exponents equal gives

8a - 2 = 12

, which simplifies to

8a = 14

, or

a = \frac{7}{4}

Rewrite each base in the equation

4^{a - 1} \cdot 8^{2a} = 16^3

as a power of 2.

(2^2)^{a-1} \cdot (2^3)^{2a} = (2^4)^3

To solve an exponential equation with different bases, rewrite the bases so they are identical.

Apply the power of a power rule

(x^m)^n = x^{m \cdot n}

to simplify each term.

2^{2a-2} \cdot 2^{6a} = 2^{12}

This simplifies the exponents by multiplying the inner and outer exponents.

Apply the product rule

x^m \cdot x^n = x^{m+n}

to combine the terms on the left side.

2^{8a-2} = 2^{12}

This combines the exponents of the terms with the common base of 2.

Set the exponents equal to each other and solve the resulting linear equation for

a

8a - 2 = 12 \Rightarrow 8a = 14 \Rightarrow a = \\frac{7}{4}

Since the bases are equal, their exponents must be equal.

Solving exponential equations by expressing all terms with a common base and applying exponent rules.

Instead of converting to base 2, all terms can be written in base 4:

4^{a - 1} \cdot (4^{1.5})^{2a} = (4^2)^3

, which simplifies to

4^{a - 1 + 3a} = 4^6

, leading to

4a - 1 = 6

and

a = \\frac{7}{4}

Tahmini Süre:1m 30s

Soru