How do you solve these logarithmic equations?

Question

How do you solve these logarithmic equations?

$4 + {log}_{9} (3 x - 7) = 6$ $4+log_(9)(3x-7)=6$
${log}_{2} (2 x) + {log}_{2} x = 5$ $log_(2)(2x)+log_(2)x=5$
$3 {log}_{5} x - {log}_{5} (5 x) = 3 - {log}_{5} 25$ $3log_(5)x-log_(5)(5x)=3-log_(5)25$

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Answer 1 · 2016-11-25T20:12:57

See below.

Explanation:

Before the logarithm application, the equations read

$9^{4} (3 x - 7) = 9^{6}$ $9^4(3x-7)=9^6$
$(2 x) x = 2^{5}$ $(2x)x=2^5$
$\frac{x^{3}}{5 x} = \frac{5^{3}}{25}$ $x^3/(5x)=5^3/25$

after that we have

$3 x - 7 = 9^{2} \to x = \frac{9^{2} + 7}{3}$ $3x-7=9^2->x=(9^2+7)/3$
$x^{2} = 2^{4} \to x = \pm 2^{2}$ $x^2=2^4->x=pm2^2$
$x^{2} = 5^{2} \to x = \pm 5$ $x^2=5^2->x=pm 5$

Answer 2 · 2016-11-25T20:18:16

$4 + {log}_{9} (3 x - 7) = 6$ $4+log_9(3x-7)=6$

${log}_{9} (3 x - 7) = 2$ $log_9(3x-7)=2$

$3 x - 7 = 81$ $3x-7=81$

$3 x = 89$ $3x=89$

$x = \frac{89}{3}$ $x=89/3$

${log}_{2} (2 x) + {log}_{2} (x) = 5$ $log_2(2x)+log_2(x)=5$

${log}_{2} (2 x^{2}) = 5$ $log_2(2x^2)=5$

$2 x^{2} = 2^{5} = 32$ $2x^2=2^5=32$

$x^{2} = 16$ $x^2=16$

$x = \pm 4$ $x=+-4$

$3 {log}_{5} x - {log}_{5} (5 x) = 3 - {log}_{5} 25$ $3log_(5)x-log_(5)(5x)=3-log_(5)25$

${log}_{5} x^{3} - {log}_{5} (5 x) = 3 - 2 = 1$ $log_5x^3-log_5(5x)=3-2=1$

${log}_{5} (\frac{x^{3}}{5 x}) = {log}_{5} (\frac{x^{2}}{5}) = 1$ $log_5(x^3/(5x))=log_5(x^2/5)=1$

$\frac{x^{2}}{5} = 5$ $x^2/5=5$

$x^{2} = 25$ $x^2=25$

$x = \pm 5$ $x=+-5$

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How do you solve these logarithmic equations?

$4 + {log}_{9} (3 x - 7) = 6$ $4+log_(9)(3x-7)=6$
${log}_{2} (2 x) + {log}_{2} x = 5$ $log_(2)(2x)+log_(2)x=5$
$3 {log}_{5} x - {log}_{5} (5 x) = 3 - {log}_{5} 25$ $3log_(5)x-log_(5)(5x)=3-log_(5)25$

2 Answers

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve these logarithmic equations?

4+log_(9)(3x-7)=64+log9(3x−7)=64+log_(9)(3x-7)=6 log_(2)(2x)+log_(2)x=5log2(2x)+log2x=5log_(2)(2x)+log_(2)x=5 3log_(5)x-log_(5)(5x)=3-log_(5)253log5x−log5(5x)=3−log5253log_(5)x-log_(5)(5x)=3-log_(5)25

2 Answers

Explanation:

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Impact of this question

$4 + {log}_{9} (3 x - 7) = 6$ $4+log_(9)(3x-7)=6$
${log}_{2} (2 x) + {log}_{2} x = 5$ $log_(2)(2x)+log_(2)x=5$
$3 {log}_{5} x - {log}_{5} (5 x) = 3 - {log}_{5} 25$ $3log_(5)x-log_(5)(5x)=3-log_(5)25$