How do you solve $3 {log}_{5} x - {log}_{5} 4 = {log}_{5} 16$ $3 log_5 x - log_5 4 = log_5 16$ ?

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Answer 1 · 2015-06-18T21:39:13

The answer is $x = 4$ $x=4$

To solve this equation you have to use the facts that:

$a \cdot {log}_{b} (c) = {log}_{b} (c^{a})$ $a*log_b(c)=log_b(c^a)$
${log}_{a} (b) - {log}_{a} (c) = {log}_{a} (\frac{b}{c})$ $log_a(b)-log_a(c)=log_a (b/c)$

First you use (1) to get:

${log}_{5} (x^{3}) - {log}_{5} (4) = {log}_{5} (16)$ $log_5(x^3) -log_5(4)=log_5(16)$

Then you use (2) to get:

${log}_{5} (\frac{x^{3}}{4}) = {log}_{5} (16)$ $log_5(x^3/4) =log_5(16)$

Now you can leave the logarithms (they both have the same base)

$\frac{x^{3}}{4} = 16$ $x^3/4=16$

$x^{3} = 64$ $x^3=64$

$x = 4$ $x=4$ (because $4^{3} = 4 \cdot 4 \cdot 4 = 64$ $4^3=4*4*4=64$ )

How do you solve 3log5x−log54=log5163 log_5 x - log_5 4 = log_5 16?