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WK14ExponentialEquationsPart6: Question 3
3). Given that
16^(-3k-2)=64^(4k+3)256^(2k+3)64^(2k-3), what is the value of
k ?

WK $14$ ExponentialEquationsPart $6$ : Question $3$ $\newline$ $3$ ). Given that $16^{-3 k-2}=64^{4 k+3} 256^{2 k+3} 64^{2 k-3}$ , what is the value of $k$ ?

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Composition of linear and quadratic functions: find a value

Full solution

Q. WK

14

ExponentialEquationsPart

6

: Question

3

\newline

3

). Given that

16^{-3 k-2}=64^{4 k+3} 256^{2 k+3} 64^{2 k-3}

, what is the value of

k

?

Simplify Exponents: First, let's simplify the equation by using the property of exponents that states $a^{(m+n)} = a^m \cdot a^n$ . $\newline$ $16^{(-3k-2)} = 64^{(4k+3)} \cdot 256^{(2k+3)} \cdot 64^{(2k-3)}$
Express Bases as Powers: Now, express all bases as powers of $2$ because $16 = 2^4$ , $64 = 2^6$ , and $256 = 2^8$ . $(2^4)^{-3k-2} = (2^6)^{4k+3} \times (2^8)^{2k+3} \times (2^6)^{2k-3}$
Apply Power of a Power Rule: Apply the power of a power rule, which is $(a^m)^n = a^{m*n}$ . $\newline$ $2^{4*(-3k-2)} = 2^{6*(4k+3)} * 2^{8*(2k+3)} * 2^{6*(2k-3)}$
Multiply Exponents: Multiply the exponents inside the parentheses. $\newline$ $2^{(-12k-8)} = 2^{(24k+18)} \times 2^{(16k+24)} \times 2^{(12k-18)}$
Set Exponents Equal: Since the bases are the same, we can set the exponents equal to each other. $\newline$ $-12k - 8 = 24k + 18 + 16k + 24 + 12k - 18$
Combine Like Terms: Combine like terms on the right side of the equation. $-12k - 8 = 52k + 24$
Isolate Variable: Add $12k$ to both sides and subtract $24$ from both sides to isolate the variable $k$ . $\newline$ $-12k + 12k - 8 + 24 = 52k + 12k + 24 - 24$
Simplify Equation: Simplify the equation. $16 = 64k$
Divide to Solve: Divide both sides by $64$ to solve for $k$ . $\newline$ $k = \frac{16}{64}$
Final Solution: Simplify the fraction. $k = \frac{1}{4}$

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