How do you solve ${log}_{3} 9^{x} = 10$ $\log _ { 3} 9^ { x } = 10$ ?

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How do you solve ${log}_{3} 9^{x} = 10$ $\log _ { 3} 9^ { x } = 10$ ?

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Answer 1 · 2017-10-15T14:58:10

$x = 5$ $x=5$

Explanation:

Using the rule ${log}_{a} M = N \Rightarrow M = a^{N}$ $log_aM=N rArr M=a^N$ you can rewrite the equation as $9^{x} = 3^{10}$ $9^x=3^10$

$9^{x} = 59049$ $9^x=59 049$
Log both sides to bring $x$ $x$ down so you can solve for it.
${log 9}^{x} = log 59049$ $log 9^x=log59049$
$x = \frac{log 59049}{log 9}$ $x=log59049/log9$
$x = 5$ $x=5$

Answer 2 · 2017-10-15T15:09:57

The answer is $5$ $5$ . See explanation.

Explanation:

${log}_{3} 9^{x} = 10$ $log_3 9^x=10$

First I use the rule:

${log}_{a} b^{c} = c \cdot {log}_{a} b$ $log_a b^c=c*log_a b$

After using this rule I get:

$x \cdot {log}_{3} 9 = 10$ $x*log_3 9=10$

${log}_{3} 9 = 2$ $log_3 9=2$ , so we get:

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How do you solve ${log}_{3} 9^{x} = 10$ $\log _ { 3} 9^ { x } = 10$ ?

2 Answers

Explanation:

Explanation:

${log}_{3} 9^{x} = 10$ $log_3 9^x=10$

${log}_{a} b^{c} = c \cdot {log}_{a} b$ $log_a b^c=c*log_a b$

$x \cdot {log}_{3} 9 = 10$ $x*log_3 9=10$

$2 x = 10 \Rightarrow x = 5$ $2x=10 => x=5$

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How do you solve log39x=10\log _ { 3} 9^ { x } = 10?

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Explanation:

Explanation:

log39x=10log_3 9^x=10

logabc=c⋅logablog_a b^c=c*log_a b

x⋅log39=10x*log_3 9=10

2x=10⇒x=52x=10 => x=5

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Impact of this question

How do you solve ${log}_{3} 9^{x} = 10$ $\log _ { 3} 9^ { x } = 10$ ?

${log}_{3} 9^{x} = 10$ $log_3 9^x=10$

${log}_{a} b^{c} = c \cdot {log}_{a} b$ $log_a b^c=c*log_a b$

$x \cdot {log}_{3} 9 = 10$ $x*log_3 9=10$

$2 x = 10 \Rightarrow x = 5$ $2x=10 => x=5$