How do you evaluate ${log}_{343} (7)$ $log_343 (7)$ ?

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How do you evaluate ${log}_{343} (7)$ $log_343 (7)$ ?

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Answer 1 · 2016-09-23T02:54:21

${log}_{343} (7) = \frac{1}{3}$ $log_343(7)=1/3$

Explanation:

${log}_{343} (7) = x$ $log_343(7)=x$

Step 1: Rewrite as an exponential.
$343^{x} = 7$ $343^x=7$

Step 2:Take the log of both sides.
${log 343}^{x} = log 7$ $log343^x=log7$

Step 3: Use the log rule ${log a}^{x} = x log a$ $loga^x=xloga$ .
$x log 343 = log 7$ $xlog343=log7$

$x = \frac{log 7}{log 343} a a a a$ $x=log7/log343color(white)(aaaa)$ Use a calculator
$x = \frac{1}{3}$ $x=1/3$

OR

Use the change of base formula ${log}_{b} a = \frac{log a}{log b}$ $log_ba=loga/logb$
${log}_{343} (7) = \frac{log 7}{log 343} = \frac{1}{3} a a a$ $log_343(7)=log7/log343=1/3color(white)(aaa)$ Use a calculator

OR

Rewrite as an exponential and consider the powers of 7.
$343^{x} = 7$ $343^x=7$

$343 = 7^{3}$ $343=7^3$ or $\sqrt[3]{343} = 7$ $root(3)343=7$ or $343^{\frac{1}{3}} = 7$ $343^(1/3)=7$
$x = \frac{1}{3}$ $x=1/3$

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How do you evaluate ${log}_{343} (7)$ $log_343 (7)$ ?

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