Solve for x using properties of logarithms: log2(32)-log3(x)=log5(25)?

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Solve for x using properties of logarithms: log2(32)-log3(x)=log5(25)?

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Answer 1 · 2018-02-22T11:31:06

$x = 1$ $x=1$

Explanation:

Since $2^{5} = 3 xx \leftrightarrow xx {log}_{2} (32) = 5$ $2^5=3color(white)("xx")harrcolor(white)("xx")log_2(32)=5$
and $x 5^{2} = 25 xx \leftrightarrow xx {log}_{5} (25) = 2$ $color(white)("x")5^2=25color(white)("xx")harrcolor(white)("xx")log_5(25)=2$

${log}_{2} (32) - {log}_{3} (x) = {log}_{5} (25)$ $log_2(32)-log_3(x)=log_5(25)$
is equivalent to
$5 - {log}_{3} (x) = 2$ $5-log_3(x)=2$

$\Rightarrow {log}_{3} (x) = 3$ $rArr log_3(x)=3$
and
since $3^{1} = 3$ $3^color(blue)1=3$
$XXX {log}_{3} (x) = 3 xx \to xx x = 1$ $color(white)("XXX")log_3(x)=3color(white)("xx")rarrcolor(white)("xx")x=color(blue)1$

Answer 2 · 2018-02-22T21:17:00

$x = 27$ $x=27$

Explanation:

${log}_{2} (32) - {log}_{3} (x) = {log}_{5} (25)$ $log_2(32)-log_3(x)=log_5(25)$

${log}_{2} (2^{5}) - {log}_{3} (x) = {log}_{5} (5^{2})$ $log_2(2^5)-log_3(x)=log_5(5^2)$

$5 {log}_{2} (2) - {log}_{3} (x) = 2 {log}_{5} (5)$ $5log_2(2)-log_3(x)=2log_5(5)$

$5 - {log}_{3} (x) = 2$ $5-log_3(x)=2$

${log}_{3} (x) = 3$ $log_3(x)=3$

Hence $x = 3^{3} = 27$ $x=3^3=27$

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Solve for x using properties of logarithms: log2(32)-log3(x)=log5(25)?

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