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Math  Logarithmic
Solve: 2^(5x6) = 7 My work: log^(5x6) = log7 5x  6(log2) = log7 5x = log7 + 6(log2) x = (log7 + log2^6) / 5 And textbook answer: (log7) / (log2) What did I do wrong? 
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I'm working on logarithmic equations and I'm stuck on how my book arrives at the next step. First, they use the change of base formula on, log(sqrt(2))(x^3  2) (sqrt(2)) is the base,changing to base 2 log(sqrt(2))(x^3  2)= log2(x^3 … 
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1.Use the laws of logarithms to express log2 (6) â€“ log2 (3) + 2log2 (8)^1/2 as a single logarithm; then evaluate. 2. For logy = log(0.5x â€“ 3) + log2, express y as a function of x. 
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solve the equation log2(x+4)log4x=2 the 2 and 4 are lower than the g This is what I got: log2(x+4)+log2(4^x)=2 log2((x+4)*4^x)=2 4^x(x+4)=4 x=0 is a solution? 
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Given that log2 3=h and log2 5=k, express log2 0.45 in terms of h and k. 
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Hello! Could someone please take a look at the problem below and let me know if I made mistakes in simplifying the given equation? 
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I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive … 
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I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive … 
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I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive … 
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i need help with these two homework problems Use the Laws of Logarithms to combine the expression into a single logarithm log2 5 − 5 log2 x + 1/2 log2(x + 1) Solve the logarithmic equation for x log2(x + 2) + log2(x − 1) …