Solve \(\log_5 (8x) + 2 \log_5 x = 3\)
Find the integer n such that n <= x < n + 1.
The first problem: log5(8x) + 2log5(x) = 3
---> log5(8x) + log5(x2) = 3 (property of logs: a multiplier goes in as an exponent)
---> log5(8x ·x2) = 3 (property of logs: adding logs = multiplying numbers)
---> log5(8x3) = 3 (simplify)
---> 8x3 = 53 (rewriting in exponential form)
---> x3 = 53/8 (divide)
---> x3 = 53/23 (rewrite 8 as 23)
---> x = 5/2 (find the cube root)