Here are some questions to go with the post on solving log equations. Answers and explanations are down at the bottom.

Find the exact value of x in each question: 2) log_{5}x – log_{5}2 = 3 2) log_{2}x – 2 = log_{2}3 3) log_{64}(x + 1) = log_{64}x + 1/2 4) 2log_{9}(x – 3) = 4 5) log_{8}(x + 1) = 1/3 – log_{8}x

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1) log_{5}x – log_{5}2 = 3 log_{5}(x/2) = 3 x/2 = 5^{3}= 125 x = 250 2) log_{2}x – 2 = log_{2}3 log_{2}x – log_{2}3 = 2 log_{2}(x/3) = 2 x/3 = 2^{2}= 4 x = 12 3) log_{64}(x + 1) = log_{64}x + 1/2 log_{64}(x + 1) – log_{64}x = 1/2 log_{64}((x + 1)/x) = 1/2 (x + 1)/x = 64^{1/2}= 8 x + 1 = 8x 7x = 1 x = 1/7 NOTE: "Exact answer" means leave your answer as a fraction unless it's an exact decimal. So 0.142857 would not be right. 4) 2log_{9}(x – 3) = 4 log_{9}(x – 3) = 2 (x – 3) = 9^{2}= 81 x = 84 ALTERNATIVE METHOD: 4) 2log_{9}(x – 3) = 4 log_{9}(x – 3)^{2}= 4 (x – 3)^{2}= 9^{4}(x – 3) = + or – 9^{2}= +81 or –81 x = 84 or~~x = –78~~x = 84 because you can't have the log of a negative number 5) log_{8}(x + 1) = 1/3 – log_{8}x log_{8}(x + 1) + log_{8}x = 1/3 log_{8}((x + 1)x) = 1/3 (x + 1)x = 8^{1/3}= 2 x^{2}+ x = 2 x^{2}+ x - 2 = 0 (x + 2)(x – 1) = 0~~x = –2~~or x = 1 x = 1 because you can't have the log of a negative number