Question 170375This question is from textbook Precalculus
: Kindly explain how to resolve the following logarithmic equation
1) 1 - log (x - 2) = log (3x + 1)
Thanks so much!!! This question is from textbook Precalculus
You can put this solution on YOUR website! Solve for x: Add to both sides. Apply the "product" rule for logarithms: Log(M)+Log(N) = Log(M*N) Perform the indicated multiplication. Recall the definition of the logarithm of a number..."The logarithm of a number is the power to which the base (base = 10 in this case) must be raised to equal that number", so... Subtract 10 from both sides. Solve by factoring. Apply the zero product rule. or so... then or
You can verify these solutions by substituting the values of x into the original equation. Notice that when subsititute x = 3, you'll get real numbers for the logarithms, but when you substitute x = -4/3, you'll get complex numbers but the two sides will be equal in both cases.
You can put this solution on YOUR website! Kindly explain how to resolve the following logarithmic equation
1) 1 - log (x - 2) = log (3x + 1)
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1 - log(x - 2) = log(3x + 1)
1 = log(3x+1)-log(x-2)
1 = log((3x+1)/(x-2)) Subtracting a log --> dividing by
Raise 10 to both sides
10 = (3x+1)/(x-2)
10*(x-2) = 3x+1
10x-20 = 3x+1
7x = 21
x = 3
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