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How do you evaluate $3 {log}_{11} 5 + {log}_{11} 7$ $3log_11 5+log_11 7$ ?

1 Answer

Apr 8, 2016

Simplified as ${log}_{11} (\frac{125}{7})$ $log_11(125/7)$
...beyond that a calculator is required to obtain: $\approx 1.202$ $~~1.202$

Explanation:

Some relations to remember:
$XXX p \cdot {log}_{b} q = {log}_{b} q^{p}$ $color(white)("XXX")p*log_b q= log_b q^p$

$XXX {log}_{b} s - {log}_{b} t = \frac{{log}_{b} s}{t}$ $color(white)("XXX")log_b s - log_b t = log_b s/t$

$3 {log}_{11} 5 XXX = {log}_{11} 5^{3} = {log}_{11} 125$ $3log_11 5color(white)("XXX")=log_11 5^3 = log_11 125$

Therefore
$3 {log}_{11} 5 - {log}_{11} 7$ $3log_11 5 - log_11 7$
$XXX = {log}_{11} 125 - {log}_{11} 7$ $color(white)("XXX")=log_11 125 - log_11 7$

$XXX = {log}_{11} (\frac{125}{7})$ $color(white)("XXX")=log_11 (125/7)$

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