Question Video: Solving Logarithmic Equations over the Set of Real Numbers Mathematics • 10th Grade

Solve logβ π₯ = logβ (1/49), where π₯ β β.

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Video Transcript

Solve log base two of π₯ is equal to log base seven of one over 49, where π₯ is a real number.

In order to answer this question, we will use our laws of logarithms. We begin by recalling that if π¦ is equal to log base π of π, then π is equal to π to the power of π¦. If we let the right-hand side of our equation log base seven of one over 49 equal π¦, then seven to the power of π¦ must equal one over 49. We know that seven squared is 49. And since π₯ to the power of negative π is equal to one over π₯ to the power of π, then seven to the power of negative two is equal to one over 49. This means that π¦ 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 π₯ equals negative two. Using the general rule stated as logarithmic functions are the inverse of exponential functions, then π₯ 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, π₯ is equal to one-quarter. The solution of the equation log base two of π₯ equals log base seven of one over 49 is π₯ equals one-quarter.