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The title of the question Solving $(2n)^{\log 2}=(5n)^{\log 5}$ asks how to solve (in some western conventions) base 10 for $(2n)^{\log 2}=(5n)^{\log 5}$. However that question's title does not seem to indicate an arbitrary base. For instance, one convention for "$\log$" is $\log c = \log_{10} c$.

Also the base $\log_b$ is considered there in the text. (See the referenced quote below).

How can the result for $\log_{10}$ be generalized for any $\log_b$? Or what are the restrictions on $b$ for it to hold?

Already, algebraically it is from one of the answers that Symbolically:

$$ \boxed{ n = \left( \frac{\displaystyle \left( 5 \right)^{\log_b 5}} { \displaystyle \left(2 \right)^{\log_b 2}} \right) ^{ \frac{\displaystyle 1} { \displaystyle \log_b 2 - \log_b 5 } } }\tag{Eq. 5}$$

But is it true that the key identity for the reduction that $n=\frac{1}{10}$ namely the identity $\left(\log_b a\right)^c = c \left(\log_b a \right)$ also holds for arbitrary numbers $b$? Or what are the restrictions for $b$?

If so, does that directly lead to the desired answer that:

$$ n= \left( \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \right) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log_b 2 -\log_b 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) $$ $$ \tag{Eq. 6a} $$

$$ n=\Bigl( 1 \Bigr) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log_b 2 -\log_b 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right)=\frac{\displaystyle 1}{\displaystyle 10} $$ $$ \tag{Eq. 6b} $$

As one claim in the answer for that question is, "This calculation is valid no matter what the base of the logarithm is." But how can the algebra be simplified step-by-step to further show this is true, answering the questions here or otherwise taking a more direct route?

Stephen Elliott
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    Raise both sides to the $\log(b)$ power to turn it into the original equation – Тyma Gaidash Mar 07 '24 at 15:04
  • Answer hint: if $(2)^{\log_b 5}=(5)^{\log_b 2}$ then the term at left is $1^{\frac{1}{log_b 2 - log_b 5}}=1$. Consider the identity $\log_b \left( \left( a^c \right) \right) = c \left( \log_b a \right)$. Applying $\log_b$ to $(2)^{\log_b 5}$ is $\log_b ((2)^{\log_b 5})=\log_b(5)\log_b(2)$. Similarly $\log_b$ to $(5)^{\log_b 2}$ is $\log_b (5^{\log_b 2})=\log_b(2)\log_b(5)$ and $(\log_b (5))(\log_b (2))=(\log_b (2))(\log_b (5))$; so indeed $(2)^{\log_b 5}=(5)^{\log_b 2}$ and $\frac{1}{(2)^{\log_b 5}(2)^{\log_b 5}}=1$ indeed directly results in Equation 6b and the desired solution. – Stephen Elliott Mar 11 '24 at 16:32

1 Answers1

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For the Questions's Equation 6a, there is a left term as follows:

$$ n= \left( \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \right)*... \tag{Question Eq. 6a}$$ And inside this left term is the quantity: $$n_{1 L}= \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right) \tag{Answer Eq. 1} $$ And this quantity can be simplified as:

$$n_{1 L}= \left(\frac{\displaystyle (5)^{\log_b 2} } {\displaystyle (2)^{\log_b 5}} \right) \tag{Answer Eq. 2} $$

Consider from the properties of logarithms (of arbitrary base, together with the restrictions on which numbers qualify) from this liked reference:

properties of logorithims

$$\tag{Reference Eqs. 1}$$

Since $2>0$ and $5>0$ are Real, it is possible to apply $\log_b$ at the left in the term and the term in the denominator using the following identity from that reference: $$ \log_b \left(x^p \right)=p \log_b \left( x \right) \tag{Answer Eq. 3} $$ It is to be tested whether $n_{1 L}=1$. To test this hypothesis, set $n_{1 L}=1 $ and continue the necessary test by simplification as:

$$ n_{1 L}=1 \text{ implies } 5^{\left( \log_b 2 \right)}=2^{\left( \log_b 5 \right)} \text{ implies } \log_b \left( 5^{\left( \log_b 2 \right)} \right) =\log_b \left( 2^{\left( \log_b 5 \right)} \right) $$ $$ \text{ implies } \left( \log_b 2 \right)\left( \log_b 5 \right) =\left( \log_b 5 \right)\left( \log_b 2 \right) \tag{Answer Eqs. 4}$$ By inspection, $$\left( \log_b 2 \right)\left( \log_b 5 \right) =\left( \log_b 5 \right)\left( \log_b 2 \right) \tag{Answer Eq. 5}$$ So clearly: $$ n_{1 L}=1 \tag{Answer Eq. 6} $$ From Equation 1, there is now the solution: $$n_{1 L}= \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right)=1 \tag{Answer Eq. 7} $$ From Question Equation 6a there is now the following simplification:

$$ n= \left( \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \right)*... \tag{Question Eq. 6a}$$ $$ n= \left( \left(1 \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \right)*... \tag{Question Eq. 6a Simplified}$$ It is simplified since $1$ raised to any real power is $1^r=1$. $$ n= \left( \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log_b 5} (5)^{-\log_b 2}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \right) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log_b 2 -\log_b 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) $$ $$ \tag{Question Eq. 6a} $$ $$ n= \left( 1 \right) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log_b 2 -\log_b 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log_b 2-\log_b 5} \right) \tag{Answer Eq. 7}$$

With the following exponential properties from this reference link and also with the following identity:

basic log_b and inverse log_b properties

$$\tag{Reference Eqs. 2}$$

$$ \left(\left(\frac{\displaystyle 1}{\displaystyle 5*2}\right) ^{\left(\log_b 2 -\log_b 5\right)}\right) ^{\frac{\displaystyle 1} {\displaystyle \left(\log_b 2 -\log_b 5\right)}} =\frac{\displaystyle 1}{10} \tag{Answer Eq. 8} $$

The above referenced identity that $y = \log_b x \Leftrightarrow b^y=x $ basically states that $\log_b f$ and $b^f$ are inverse operations. Also, from the same reference, $\log_b \left( c^d \right) = d \log_b c$ and the Answer Equation 8 was applied as follows:

$$ n=b^\left( {\displaystyle \log_b} \left( \left( \left(\frac{\displaystyle 1}{\displaystyle 5*2} \right)^{\displaystyle \left(\log_b 2 -\log_b 5\right)} \right)^{\frac{\displaystyle 1} {\displaystyle \left(\log_b 2 -\log_b 5\right)}} \right) \right) $$ $$ n=b^\left( \left( \left( {\frac{\displaystyle 1} {\displaystyle \left(\log_b 2 -\log_b 5\right)}} \right) {\displaystyle \log_b} \left( \left(\frac{\displaystyle 1}{\displaystyle 5*2} \right)^{\displaystyle \left(\log_b 2 -\log_b 5\right)} \right) \right) \right) $$ $$ n=b^\left( \left( \left( \displaystyle \log_b 2 -\log_b 5 \right) \left( {\frac{\displaystyle 1} {\displaystyle \left(\log_b 2 -\log_b 5\right)}} \right) {\displaystyle \log_b} \left( \left(\frac{\displaystyle 1}{\displaystyle 5*2} \right) \right) \right) \right) $$ $$ n=b^\left( \left( {\displaystyle 1 *} {\displaystyle \log_b} \left( \frac{\displaystyle 1}{\displaystyle 5*2} \right) \right) \right) \text{ since } \left( \displaystyle \log_b 2 -\log_b 5 \right) \left( {\frac{\displaystyle 1} {\displaystyle \left(\log_b 2 -\log_b 5\right)}} \right)={\displaystyle 1} $$ $$\tag{Answer Eqs. 9} $$

Now from the Mathematics Stack Exchange Article, Proof that $ b^{\log_b(x)} = x$ there is the direct relationship that $b^{\log_b(x)}=x$. Thus applying that identity to Equations 9:

$$ \boxed{ n=b^\left( \left( {\displaystyle 1 *} {\displaystyle \log_b} \left( \frac{\displaystyle 1}{\displaystyle 5*2} \right) \right) \right) = b^\left( \left( {\displaystyle \log_b} \left( \frac{\displaystyle 1}{\displaystyle 5*2} \right) \right) \right) =\frac{\displaystyle 1}{\displaystyle 5*2} \text{ so finally } n = \frac{\displaystyle 1}{\displaystyle 10} } \tag{Eq. 10} $$

Stephen Elliott
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