Solve each equation. log￬1/2 (x+3) = -4

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Accepted Answer

Hey, everyone in this problem, we're asked to solve the following logarithmic equation for X. We have log base 1/5 of X plus 98 is equal to negative three. We're given four answer choices. Option A is the set containing nine. Option B is the set containing negative 18. Option C is a set containing 27. And option D is a set containing negative 21. And we're gonna start by rewriting the expression. We were given the long base 1/ of X plus 98. It is equal to negative three. Now our X is inside of this log. So in order to solve for X, we need to get it out of this lock. Now recall that we have a rule that relates log equations to exponentials. OK. So if we have log base B of A is equal to Y, we can rewrite this as B to the exponent Y is equal to A and so the base of the log becomes the base of the exponent. Now, in our case, the base of our log is 1/5. So we're gonna have 1/5 the base of our exponent. The value on the opposite side of the equation to the log negative three is our Y value that becomes the exponent, the 1/5 to the exponent negative three is equal to A which is everything inside of the log in our case X plus 98. Now, on the left hand side, we have 1/5 to the exponent negative three. When we have a fraction raised to an exponent. Can you recall that, that exponent applies to both the numerator and the denominator individually? Now the numerator is just 11 raise to the exponent of negative three is going to be just one. So in the numerator, we have one and in the denominator, we have five to the exponent negative three and this is equal to X plus 98. Now we have a negative exponent in the denominator. How can we, we write this with a positive exponent? Can we know how to evaluate positive exponents? How can we write this as a positive exponent? We recall if we have one divided by a, to the exponent negative M, we can write this as a to the exponent positive app. OK. So if we have a negative exponent in the denominator, we can move the base and the exponent to the numerator and that negative exponent becomes positive. OK. So the left hand side of our equation one divided by five to the exponent negative three becomes five cubed. And this is equal to X plus 98. Now five cubed, we know that this is equal to 125. So we have 125 is equal to X plus 98. Subtracting 98 from both sides to isolate for X, we get that X is equal to 27. And so the correct answer choice here is going to be C we have the set containing 27. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each equation. log￬1/2 (x+3) = -4

Key Concepts

Logarithmic Functions

Change of Base Formula

Exponential Equations

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