algebra-precalculusexponential functionlogarithmsratio
Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?
Edit: @mathlove noticed problems with my initial solution. My effort to resolve those problems ended in a major revision of the solution. Please comment if you find mistakes in this new version.
Let's take off from where you calculated that $t^2-2t=a-\sqrt{a+t}$ . Our aim is to find a value (or values) for $a$ such that the above equation has exactly 2 real answers for $t$. Let's also keep in mind the implicit constraint that $$a+t \ge 0 \Rightarrow t \ge -a \qquad(1)$$ This is because the given problem includes $\sqrt{a+t}$ . An important implication of (1) is that if $a$ is negative then $t$ should be positive, and if $t$ is negative then $a$ should be positive.
Another constraint that is derived from our starting equation is that $$t^2-2t \le a \qquad (2)$$ where the equality holds when $a+t=0$. Besides, (2) means that if $a < 0$ then $t^2-2t=t(t-2) < 0$ , which implies that $0 < t < 2$.
And two other constraints, which we can derive from (2) are $$t^2-2t+1 = (t-1)^2 \le a+1 \qquad (3)$$ $$0 \le (t-1)^2 \le a+1 \Longrightarrow a \ge -1 \qquad (4)$$
Alright, here we go: $$t^2-2t=a-\sqrt{a+t}$$ $$t^2-t=a+t-\sqrt{a+t}$$ $$t^2-t+\frac{1}{4}=a+t-\sqrt{a+t}+\frac{1}{4}$$ $$(t-\frac{1}{2})^2 = (\sqrt{a+t}-\frac{1}{2})^2$$ $$t-\frac{1}{2} = \pm(\sqrt{a+t}-\frac{1}{2})$$ So we have two (quadratic) equations $$t-\frac{1}{2} = +(\sqrt{a+t}-\frac{1}{2}) \Rightarrow t^2 = a+t \qquad (5)$$ and $$t-\frac{1}{2} = -(\sqrt{a+t}-\frac{1}{2}) \Rightarrow (t-1)^2 = a+t \qquad (6)$$ Let's look more carefully at (5). The solutions to this quadratic equation are $$t_1 = \frac{1 + \sqrt{1+4a}}{2} , t_2 = \frac{1 - \sqrt{1+4a}}{2}$$ For these solutions to exist, we should have $a \ge -\frac{1}{4}$.
Consider $t_1$. There seems to be no problem with $t_1$ for $-\frac{1}{4} \le a$ . In this range of $a$ , $t_1$ seems to satisfy constraints (1),(2),(3) and (4).
Now consider $t_2$. If $a > 0$ then by constraint (2) we should have: $$t_2^2 - 2t_2 \le a$$ $$(\frac{1 - \sqrt{1+4a}}{2})^2 - 2(\frac{1 - \sqrt{1+4a}}{2})= \frac{-1+2a+\sqrt{1+4a}}{2} \le a$$ $$-1+2a+\sqrt{1+4a} \le 2a$$ $$\sqrt{1+4a} \le 1$$ $$1+4a \le 1$$ $$a \le 0$$ which is a contradiction with our assumption of $a > 0$. So, $t_2$ does not exist for $a > 0$. There seems to be no problem with $t_2$ for $-\frac{1}{4} \le a \le 0$ . In this range of $a$ , $t_2$ seems to satisfy constraints (1),(2),(3) and (4).
Note that if $a = -\frac{1}{4}$ then $t_1=t_2=\frac{1}{2}$.
Let's now look at (6). The solutions to this quadratic equation are $$t_3 = \frac{3 + \sqrt{5+4a}}{2} , t_4 = \frac{3 - \sqrt{5+4a}}{2}$$ For these solutions to exist, we should have $a \ge -\frac{5}{4}$ , but we also have the stricter constraint (4), which requires that $a \ge -1$.
Consider $t_3$. If $a > 0$ then $t_3$ violates constraint (3). So, $t_3$ does not exist for $a > 0$. On the other hand, if $-1 < a \le 0$ then by constraint (1) t should be non-negative and by the implication of constraint (2) we should have $0 \le t \le 2$. But applying this range of $a$ in the calculation of $t_3$ results in $2 < t_3$, which is in contradiction with constraint (2). Finally, note that if $a=-1$ then $t_3=2$ , but this result violates constraint (3). Therefore, it seems that $t_3$ is not a solution to the equation for any value of $a$.
Now consider $t_4$. It seems that for $a \ge -1$ , $t_4$ satisfies constraints (1),(2),(3) and (4).
Note that although in the formula we get $t_3=t_4=\frac{3}{2}$ for $a=-\frac{5}{4}$ , these are not solutions of the main equation, because the value of $a=-\frac{5}{4}$ violates constraint (4).
Now let us see in what conditions $t_1$ or $t_2$ might be equal to $t_4$. From $t_1 = t_4$ it follows that $a=-\frac{1}{4}$. Also, from $t_2 = t_4$ it follows that $a=-\frac{1}{4}$. So at $a=-\frac{1}{4}$ , $t_1=t_2=t_4=\frac{1}{2}$.
Summing up our above analyses, we can see that
$t_1$ exists for $-\frac{1}{4} \le a$
$t_2$ exists for $-\frac{1}{4} \le a \le 0$
$t_3$ does not exist
$t_4$ exists for $-1 \le a$
Therefore, for $a > 0$ , the equation has exactly two distinct solutions ($t_1 , t_4$).
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.
[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.
Best Answer
It seems easier to use $\log_{12}$ rather than $\log$.
$\log_{12}(36)=\log_{12}(3)+1=k$, and $$2\log_{12}(2)+\log_{12}(3)=\log_{12}(12)=1,$$ so $2\log_{12}(2)=2-k$, or $$\log_{12}(2)=1-\frac{k}{2}$$
Thus $$\log_{24}(48)= \frac{\log_{12}(48)}{\log_{12}(24)}=\frac{2\log_{12}(2)+1}{\log_{12}(2)+1}$$ $$=\frac{2-k+1}{2-\frac{k}{2}}=\frac{6-2k}{4-k}.$$