Respuesta :
[tex]log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x[/tex]
Solution:
Given that,
[tex]log_b2 = x\\\\log_b3 = y --------(i)[/tex]
Use the following log rules
Rule 1: [tex]log_b(ac) = log_ba + log_bc[/tex]
Rule 2: [tex]log_b\frac{a}{c} = log_ba - log_bc[/tex]
Rule 3: [tex]log_ba^c = clog_ba[/tex]
[tex]a) log_b{162}[/tex]
Break 162 down to primes:
[tex]162 = 2^1 \times 3^4[/tex]
[tex]log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y[/tex]
Thus we get,
[tex]log_b162 = x + 4y[/tex]
Next
[tex]b) log_b 324[/tex]
Break 324 down to primes:
[tex]324 = 2^2 \times 3^4[/tex]
[tex]log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y[/tex]
Next
[tex]c) log_b\frac{8}{9}[/tex]
By rule 2
[tex]log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y[/tex]
Next
[tex]d) \frac{log_b27}{log_b16}[/tex]
By rule 2
[tex]\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x[/tex]
Thus the given are evaluated in terms of x and y
