Help pls!!! With explanations
Help pls!!! With explanations
\(\frac{log_{2}(24)}{log_{96}(2)} - \frac{log_{2}(192)}{log_{12}(2)}\)
log₂(24): \(2^{x} = 24\)
\(ln(2^{x}) = ln(24)\)
\(xln(2) = ln(24)\)
\(x = \frac{ln(24)}{ln(2)}\)
\(x = \frac{ln(8 * 3)}{ln(2)}\)
\(x = \frac{ln(8) + ln(3)}{ln(2)}\)
\(x = \frac{3ln(2) + ln(3)}{ln(2)}\)
\(x = \frac{3ln(2)}{ln(2)} + \frac{ln(3)}{ln(2)}\)
\(x = 3 + 1.1\)
\(x = 4.1\)
log₉₆(2): \(96^{x} = 2\)
\(ln(96^{x}) = ln(2)\)
\(xln(96) = ln(2)\)
\(xln(32 * 3) = ln(2)\)
\(x(ln(32) + ln(3)) = ln(2)\)
\(\frac{x(ln(32) + ln(3))}{ln(32) + ln(3)} = \frac{ln(2)}{ln(32) + ln(3)}\)
\(x = \frac{ln(2)}{ln((2)^{5}) + ln(3)}\)
\(x = \frac{ln(2)}{5ln(2) + ln(3)}\)
\(x = 0.7\)
\(\frac{log_{2}(24)}{log_{96}(2)} = \frac{3 + \frac{ln(3)}{ln(2)}}{\frac{ln(2)}{5ln(2) + ln(3)}} = \frac{4.1}{0.7} = 5\frac{6}{7}\)
log₂(192): \(2^{x} = 192\)
\(ln(2^{x}) = ln(192)\)
\(xln(2) = ln(64 * 3)\)
\(xln(2) = ln(64) + ln(3)\)
\(xln(2) = ln((2)^{6}) + ln(3)\)
\(xln(2) = 6ln(2) + ln(3)\)
\(\frac{xln(2)}{ln(2)} = \frac{6ln(2) + ln(3)}{ln(2)}\)
\(x = \frac{6ln(2)}{ln(2)} + \frac{ln(3)}{ln(2)}\)
\(x = 6 + 1.1\)
\(x = 7.1\)
log₁₂(2): \(12^{x} = 2\)
\(ln(12^{x}) = ln(2)\)
\(xln(12) = ln(2)\)
\(xln(4 * 3) = ln(2)\)
\(x(ln(4) + ln(3)) = ln(2)\)
\(\frac{x(ln(4) + ln(3))}{ln(4) + ln(3)} = \frac{ln(2)}{ln(4) + ln(3)}\)
\(x = \frac{ln(2)}{ln((2)^{2}) + ln(3)}\)
\(x = \frac{ln(2)}{2ln(2) + ln(3)}\)
\(x = 0.3\)
\(\frac{log_{2}(192)}{log_{12}(2)} = \frac{log_{2}(192)}{log_{12}(2)} = \frac{7.1}{0.3} = 23\frac{2}{3}\)
\(5\frac{6}{7} - 23\frac{2}{3}\)
\(-17\frac{17}{21}\)