If logb2=x and logb3=y, evaluate the following in terms of x and y: (a) logb162= (b) logb324= (c) logb8/9=

Question

If logb2=x and logb3=y, evaluate the following in terms of x and y:

(a) logb162=
(b) logb324=
(c) logb8/9=
(d) logb27/logb16

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Isabella 1 week 2021-11-21T14:21:20+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-21T14:23:06+00:00

    log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

    Solution:

    Given that,

    log_b2 = x\\\\log_b3 = y --------(i)

    Use the following log rules

    Rule 1: log_b(ac) = log_ba + log_bc

    Rule 2: log_b\frac{a}{c} = log_ba - log_bc

    Rule 3: log_ba^c = clog_ba

    a) log_b{162}

    Break 162 down to primes:

    162 = 2^1 \times 3^4

    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

    Thus we get,

    log_b162 = x + 4y

    Next

    b) log_b 324

    Break 324 down to primes:

    324 = 2^2 \times 3^4

    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

    Next

    c) log_b\frac{8}{9}

    By rule 2

    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

    Next

    d) \frac{log_b27}{log_b16}

    By rule 2

    \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

    Thus the given are evaluated in terms of x and y

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