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Solve log₂ 𝑥 = log₇ (1/49), where 𝑥 ∈ ℝ.

Solve log base two of 𝑥 is equal to log base seven of one over 49, where 𝑥 is a real number.

In order to answer this question, we will use our laws of logarithms. We begin by recalling that if 𝑦 is equal to log base 𝑎 of 𝑏, then 𝑏 is equal to 𝑎 to the power of 𝑦. If we let the right-hand side of our equation log base seven of one over 49 equal 𝑦, then seven to the power of 𝑦 must equal one over 49. We know that seven squared is 49. And since 𝑥 to the power of negative 𝑛 is equal to one over 𝑥 to the power of 𝑛, then seven to the power of negative two is equal to one over 49. This means that 𝑦 is equal to negative two.

As log base seven of one over 49 equals negative two, we can rewrite our initial equation as log base two of 𝑥 equals negative two. Using the general rule stated as logarithmic functions are the inverse of exponential functions, then 𝑥 is equal to two to the power of negative two. This is equal to one over two squared. As two squared is equal to four, 𝑥 is equal to one-quarter. The solution of the equation log base two of 𝑥 equals log base seven of one over 49 is 𝑥 equals one-quarter.

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