How do you solve ${log}_{5} x + {log}_{3} x = 1$ $Log_(5)x + log_(3)x = 1$ ?

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How do you solve ${log}_{5} x + {log}_{3} x = 1$ $Log_(5)x + log_(3)x = 1$ ?

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Answer 1 · 2016-02-23T03:08:47

$x = e^{\frac{ln 5 (ln 3)}{ln 3 + ln 5}} \approx 1.9211$ $x=e^((ln5(ln3))/(ln3+ln5))approx1.9211$

Explanation:

Use the change of base formula, which states that

${log}_{a} b = \frac{{log}_{c} b}{{log}_{c} a}$ $log_ab=log_cb/log_ca$

The common base I'll use here is $e$ $e$ , since it is the base of the natural logarithm. It doesn't actually matter which base is chosen: any base will work.

The original expression can be rewritten as:

$\frac{ln x}{ln 5} + \frac{ln x}{ln 3} = 1$ $lnx/ln5+lnx/ln3=1$

Find a common denominator of $ln 5 (ln 3)$ $ln5(ln3)$ :

$\frac{ln x (ln 3)}{ln 5 (ln 3)} + \frac{ln x (ln 5)}{ln 5 (ln 3)} = 1$ $(lnx(ln3))/(ln5(ln3))+(lnx(ln5))/(ln5(ln3))=1$

$\frac{ln x (ln 3) + ln x (ln 5)}{ln 5 (ln 3)} = 1$ $(lnx(ln3)+lnx(ln5))/(ln5(ln3))=1$

Cross multiply.

$ln x (ln 3) + ln x (ln 5) = ln 5 (ln 3)$ $lnx(ln3)+lnx(ln5)=ln5(ln3)$

Factor a $ln x$ $lnx$ from both terms on the left hand side.

$ln x (ln 3 + ln 5) = ln 5 (ln 3)$ $lnx(ln3+ln5)=ln5(ln3)$

Divide both sides by $ln 3 + ln 5$ $ln3+ln5$ .

$ln x = \frac{ln 5 (ln 3)}{ln 3 + ln 5}$ $lnx=(ln5(ln3))/(ln3+ln5)$

To undo the natural logarithm, exponentiate both sides with base $e$ $e$ .

$x = e^{\frac{ln 5 (ln 3)}{ln 3 + ln 5}} \approx 1.9211$ $x=e^((ln5(ln3))/(ln3+ln5))approx1.9211$

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How do you solve ${log}_{5} x + {log}_{3} x = 1$ $Log_(5)x + log_(3)x = 1$ ?

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How do you solve log5x+log3x=1Log_(5)x + log_(3)x = 1?

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How do you solve ${log}_{5} x + {log}_{3} x = 1$ $Log_(5)x + log_(3)x = 1$ ?