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\log _2(x+1)=\log _3(27) \ln (x+2)-\ln (x+1)=1 \ln (x)+\ln (x-1)=\ln (3x+12) 4+\log _3(7x)=10 \ln (10)-\ln (7-x)=\ln (x) \log _2(x^2-6x)=3+\log _2(1-x)

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Type 1. In this type, the variable you need to solve for is inside the log, with one log on one side of the equation and a constant on the other. Turn the variable inside the log into an

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Since we want to transform the left side into a single logarithmic equation, we

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The answer to the equation is 4.

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Solve the logarithmic equation: $$ \log_{2}(x+4)-\log_{2}(3)=\log_{2}(x-2)-\log_{2}(5)$$ Solution: Here, we have a logarithm subtraction on each side of the equation. We can use the quotient

log(3x−2)−log(2) = log(x+4) log(3x−2 2) = log(x+4) Apply the quotient rule of logarithms. 3x−2 2 = x+4 Apply the one-to-one property. 3x−2 = 2x+8 Multiply both sides of the equation by 2. x = 10 Subtract 2x and add 2.

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