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How does one solve ${\log _3}15$ ?

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint:For solving this particular problem we will use ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ , change of base rule may be used if $a$ and $b$ are greater than zero and not adequate to one , and $x$ is larger than zero . For simplifying the equation , we will use the logarithm property that is $\log ab = \log a + \log b$ .

Formula used:
We used logarithm property i.e., The change of base rule may be used if $a$
and $b$ are greater than zero and not adequate to one , and $x$ is larger than zero .
${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$
and $\log ab = \log a + \log b$ .

Complete solution step by step:
We have to find the value of ${\log _3}15$ ,
The change of base rule may be used if $a$
and $b$ are greater than zero and not adequate to one , and $x$ is larger than zero .
${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$
Now, substitute values for the variables within the change of base formula, using $x = 10$.
${\log _3}15 = \dfrac{{{{\log }_{10}}15}}{{{{\log }_{10}}3}}$
Using logarithm property that is $\log ab = \log a + \log b$ ,
We can write ,
$
\Rightarrow {\log _3}15 = \dfrac{{{{\log }_{10}}5 + {{\log }_{10}}3}}{{{{\log }_{10}}3}} \\
\Rightarrow {\log _3}15 = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}} + 1 \\
$
Since $\dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}} = 1.46497352$ ,
Therefore , we get the following result ,
\[
\Rightarrow {\log _3}15 = 1.46497352 + 1 \\
\Rightarrow {\log _3}15 = 2.46497352 \\
\]

The result is shown in multiple forms.
Exact Form:
${\log _3}15 = \dfrac{{{{\log }_{10}}15}}{{{{\log }_{10}}3}}$
Decimal Form:
\[{\log _3}15 = 2.46497352\]

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have remember certain conditions , our end result must satisfy domain of that logarithm