Solving $(2n)^{\log 2}=(5n)^{\log 5}$

Question

I have seen this equation from a link named Asisten and German Academy, (it is a video of Facebook) where there is a complicate solution (I invite to watch it) for

$$(2n)^{\log 2}=(5n)^{\log 5}$$

I have adopted, instead, this approach: $$(2)^{\log 2}(n)^{\log 2}=(5)^{\log 5}(n)^{\log 5} \iff 2n^{\log 2}=5n^{\log 5}$$

After

$$\frac{n^{\log 2}}{n^{\log 5}}=\frac 52 \iff n^{(\log 2-\log 5)}=\frac 52$$

$$\log(n^{(\log 2-\log 5)})= \log 5-\log 2 $$

$$(\log 2-\log 5)\log n=\log 5-\log 2 \iff \log n =-1$$

Hence $$e^{\log n}=e^{-1}\implies n=\frac 1e$$

Now I have seen the solution is $n=1/10$.

Is it different my solution why my base is $e$ and not $10$? Generally I have seen that $\log=\log_{10}$. In Italy we used often $\log=\log_e$. I yet thought with the old notation that $\operatorname{Log}=\log_{10}.$ I not adopted often $\ln$ where the base is neperian.

Answer 1

Regardless of what base you are using, it is not true that both $$ 2^{\log(2)} = 2 \qquad\text{and}\qquad 5^{\log(5)} = 5. $$ It is certainly true that $2^{\log_2(2)} = 2$, but the notation $\log$, without a subscript, does not mean "choose whatever base you like and go with it," it typically denotes a specific base.

In mathematics, $\log$ typically means the natural logarithm,^[1] which has base $\mathrm{e}$. Under that assumption, $$ 2^{\log(2)} \approx 1.617 \qquad\text{and}\qquad 5^{\log(5)} \approx 13.334. $$ If $\log$ denotes the common logarithm, which as base $10$, then $$ 2^{\log(2)} \approx 1.232 \qquad\text{and}\qquad 5^{\log(5)} \approx 3.080. $$

In any event, assuming that $\log$ is the natural logarithm, $$\begin{align} &(2n)^{\log(2)} = (5n)^{\log(5)} \\ &\qquad\iff \frac{2^{\log(2)}}{5^{\log(5)}} = \frac{n^{\log(5)}}{n^{\log(2)}} \\ &\qquad\iff\frac{\mathrm{e}^{\log(2)^2}}{\mathrm{e}^{\log(5)^2}} = \frac{n^{\log(5)}}{n^{\log(2)}} \label{eq1}\tag{*} \\ &\qquad\iff \mathrm{e}^{\log(2)^2 - \log(5)^2} = n^{\log(5)-\log(2)} \\ &\qquad\iff \log(2)^2 - \log(5)^2 = \log(n) \left( \log(5)-\log(2) \right) \\ &\qquad\iff \log(n) = \frac{\log(2)^2 - \log(5)^2}{\log(5)-\log(2)} = -\left( \log(2)+\log(5)\right) = -\log(10) \\ &\qquad\iff n = \mathrm{e}^{-\log(10)} = \frac{1}{10}. \end{align}$$

At (\ref{eq1}), I am using the fact that for any positive real number $a$, we can write $$ a = \exp(\log(a)) = \mathrm{e}^{\log(a)}, $$ since the exponential and logarithm are inverse functions to each other. This implies that, for exmaple, $$ 2^{\log(2)} = \left( \mathrm{e}^{\log(2)} \right)^{\log(2)} = \mathrm{e}^{\log(2)\cdot \log(2)} = \mathrm{e}^{\log(2)^2}. $$

Note that the choice of the natural logarithm is irrelevant here. Replacing the natural logarithm with the logarithm with base $b$ will have the effect of replacing every occurrence of $\mathrm{e}$ with $b$ in the above computation, but this will not change the result.

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[1] The question asks about "neperian" logarithms. This term does not exist in English, though it probably refers to the Napierian logarithm (also spelled "Naperian"), named for John Napier, the Scottish mathematician responsible for working out large tables of logarithm ("Napier's bones"). The Napierian logarithm is the natural logarithm.

Answer 2

We have that $a^{\log a} = e^{\log^2 a}$ then

$$(2n)^{\log 2}=(5n)^{\log 5} $$

$$2^{\log 2}n^{\log 2}=5^{\log 5}n^{\log 5}$$

$$e^{\log^2 2+\log n \log 2}=e^{\log^2 5+\log n \log 5}$$

and since $e^x$ is striclty monotonic, the latter is equivalent to

$$\log^2 2+\log n \log 2=\log^2 5+\log n \log 5$$

$$\log n = \frac{\log^2 2-\log^2 5}{\log 5 -\log 2}=-\log 10 \iff n=\frac1{10}$$

Answer 3

As already pointed out in the comments and in Xander's answer, your calculation is wrong because $2^{\log_b 2} = 2$ and $5^{\log_b 5} = 5$ do not hold simultaneously for any base $b$.

I would argue as follows: $$ \begin{align} &(2n)^{\log 2}=(5n)^{\log 5} \\ \iff &\log 2 (\log 2 + \log n) = \log 5 (\log 5 + \log n) \\ \iff &\log n = - \frac{(\log 5)^2 - (\log 2)^2}{\log 5 - \log 2} = -\log(10) = \log \frac{1}{10} \\ \iff &n = \frac{1}{10} \, . \end{align} $$

This calculation is valid no matter what the base of the logarithm is.

Answer 4

Why don't you simply take the logarithm of both sides:

$(2n)^{\log 2} = (5n)^{log 5} \iff$ $\log 2 \log(2n) = \log 5 \log(5n)$

This can be elaborated and simplified:

$\log 2 (\log 2 + \log n) = \log 5 (\log 5 + \log n)$
$\log n = -(\log 2 + \log 5) = -\log 10 \iff n=\frac{1}{10}$

Because:
$a(a+x) = b(b+x)$
$a^2-b^2 = (b-a)x$
$x=-(a+b)$

Answer 5

The question Equation 1 (is repeated here to easier reference) is:

$$ \left(2 n \right)^{\log 2} = \left( 5 n \right)^{\log 5} \tag{Eq. 1}$$ Then, $$\begin{align*} \left(2 \right)^{\log 2}\left(n \right)^{\log 2} &= \left( 5 \right)^{\log 5} \left( n \right)^{\log 5} \tag{Eq. 2}\\ \frac{\left(n \right)^{\log 2}}{ \left( n \right)^{\log 5}} &= \frac{\left( 5 \right)^{\log 5}} { \left(2 \right)^{\log 2}} \tag{Eq. 3}\\ n^{\log 2-\log 5 } &= \frac{\displaystyle \left( 5 \right)^{\log 5}} {\displaystyle \left(2 \right)^{\log 2}} \tag{Eq. 4} \end{align*}$$ Finally, the result is given in Equation 5 as: $$ \boxed{ n = \left( \frac{\displaystyle \left( 5 \right)^{\log 5}} { \displaystyle \left(2 \right)^{\log 2}} \right) ^{ \frac{\displaystyle 1} { \displaystyle \log 2 - \log 5 } } = \frac{1}{10} }\tag{Eq. 5a}$$

Also, since all of the symbolic algebraic operation was independent of the base $\log_b$, the same result holds independent of base $b$ so long that $b>0$ and $b \ne 1$ (with algebraic steps referenced in this link):

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

Simplification of Equation 5 to Get to $n=\frac{1}{10}$

From Equation 5, it can be rewritten as follows: $$ n=\left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log 2} (5)^{-\log 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right) $$ $$ n= \left( \left(\frac{\displaystyle 1 } {\displaystyle (2)^{\log 5} (5)^{-\log 2}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right) \right) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log 2 -\log 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right)$$ $$ \tag{Eq. 6} $$ For the professionals, the symbolic representation of $0.1$ is sufficient to be derived using standard symbolic math manipulation techniques. Especially the identities "$a^n*b^n=(a*b)^n$" , "$(a^n)^{\frac{1}{n}}=a$" for "$a \ge 0$", and $a^{\left(\log b \right)} = b^{\left(\log a \right)}$ are helpful in the above simplifications.

In particular, the simplifications and Equations 6 together with the identity $a^{\left(\log b\right)} = b^{\left(\log a \right)}$ imply that

$$ \left(\frac{\displaystyle (5)^{\log 2} } {\displaystyle (2)^{\log 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right) =\left({\displaystyle 1 }\right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right)=1 \tag{Eqs. 7}$$ $$ n=\left({\displaystyle 1 }\right) \left(\frac{\displaystyle 1 } {\displaystyle (2*5)^{\log 2 -\log 5}} \right)^ \left(\frac{\displaystyle 1}{\displaystyle \log 2-\log 5} \right) \tag{Eq. 8} $$

And finally, referencing the identity $\left(a^n\right)^{\frac{1}{n}}=a$ for $a>0$ and $n \ne 0$, Equation 8 leads to the desired solution.

Also, more step-to-step simplification steps are given this reference, "Algebraically Simplify $n =\left( \frac{ \left(5\right)^{\log 5}} { \left(2 \right)^{\log 2}} \right) ^{ \frac{1} {\log 2-\log 5 } } = \frac{1}{10}\text{"}$.

With both either reference link or use of the identity $\left(a^n\right)^{\frac{1}{n}}=a$ options, immediately the decimal simplification and answer from Equation 8 becomes apparent: $$ \boxed{ n= \left({\displaystyle 1}\right) \left( \frac{\displaystyle 1} {\displaystyle 2} \right) \left( \frac{\displaystyle 1} {\displaystyle 5} \right) =\frac{\displaystyle 1}{\displaystyle 10} } \tag{Eqs. 9}$$

Symbolic to Decimal Verification

This result can also easily be verified with a scientific calculator (done already below), so it is definitely correct: Also with the Wolfram Online Symbolic Calculator: The decimal result is the same, that $n=0.1$:

Answer 6

If we let $n=\frac k{10}$, then $$ \begin{align} &(2n)^{\log 2}=(5n)^{\log 5} \\ \iff &(\tfrac k5)^{\log2}=(\tfrac k2)^{\log5}\\ \iff &5^{-\log2}k^{\log2}=2^{-\log5}k^{\log5}\\ \stackrel{(*)}{\iff} &k^{\log2}=k^{\log5}\\ \iff &k^{\log5-\log2}=1\\ \iff &k=1 \;\text{and}\; n=\frac1{10} \end{align} $$

$(*)$ By taking $\log$ of both sides, $$5^{-\log2}=2^{-\log5}\iff -\log5\log2=-\log2\log5.$$