# What's wrong with my aproach to solving this equation with multiple logarithms?

A question I was faced with asked "For which $x$ is $\log_{10}(x)^{\log_{10}(\log_{10}(x))}= 10,000$?"

My instincts tell me I can say $$\log_{10}(x)=10$$ and $$\log_{10}(\log_{10}(x))=4$$

However, this leads to an incorrect answer. Instead, the solution posits I let $a=\log_{10}(x)$, and take the $\log$ base ten of both sides. Why is my answer wrong?

• Note that your solution would give the LHS as $\log_{10}(x)^{\log_{10}(10)}=\log_{10}^{\log_{10}1}=\log_{10}^0 =1$. So I'd say your instincts are simply incorrect here. – Semiclassical Jul 22 '14 at 5:58
• What you've tried is fine; it's just that it doesn't yield a solution. Otherwise, you may have found a solution that way. – Myridium Jul 22 '14 at 6:24
• Am I the only one who finds the way the question is written ambiguous? Shouldn't it be $(\log_{10}(x))^{\log_{10}(\log_{10}(x))}= 10,000$? – Deepak Jul 22 '14 at 7:02
• @Deepak I thought the brackets made it explicit enough. What would be ambiguous is e.g. $\log a^b$, which is where people either specify $\log (a^b)$, or $(\log a)^b$, or $\log(a)^b$ (or, equivalently, $\log^b a$). – Thomas Jul 22 '14 at 13:05
• @Thomas When people write $\log a^b$ I think it's pretty standard convention to assume the meaning to be $\log (a^b) = b\log a$. Of course, writing $(\log a)^b$ is completely unambiguous (which is why I preferred this). But as for this notation: $\log (a)^b$, I haven't learnt that it can be unambiguously interpreted as $(\log a)^b$. In fact, I'm more inclined to interpret it as $\log (a^b) = b\log a$. – Deepak Jul 22 '14 at 14:00

$$\log_{10}(x)=10\Rightarrow x=10^{10}$$ and $$\log_{10}(\log_{10}(x))=\log_{10}10=1\neq 4$$

Taking $\log_{10}$ of both sides and using the rule $\log_{10}(a^b)=b\log_{10}a$ gives $$(\log_{10}(\log_{10}(x))(\log_{10}(\log_{10}(x))=\log_{10}10000=4\ ,$$ that is, $$(\log_{10}(\log_{10}(x)))^2=4\ .$$ Hence $$\log_{10}(\log_{10}(x))=2\quad\Rightarrow\quad \log_{10}(x)=100\quad\Rightarrow\quad x=10^{100}$$ or $$\log_{10}(\log_{10}(x))=-2\quad\Rightarrow\quad \log_{10}(x)=\frac{1}{100}\quad\Rightarrow\quad x=10^{1/100}\ .$$

• Although a good answer, the OP asked why his answer was wrong. +1 Anyways. – Vishwa Iyer Jul 22 '14 at 11:03
• @Vishwa Fair point, though I'm not sure there really is any answer to that beyond what Semiclassical said in a comment. – David Jul 22 '14 at 14:07

There's nothing wrong with the technique you've employed; you've made an assumption (namely $\log_{10}(x)=10$) and solved alfebraically for what $x$ must be given the assumption.

$x$ having no solution here tells you that if your assumption was correct, then there is no solution to the equation. So either there's no solution to $x$ or your assumption was wrong.

The concept of imposing such assumptions and solving the outcome is, I think, a very important one to understand. Although the assumption you presented could have yielded a solution, it didn't. This doesn't mean that there is no solution at all, just that there is no solution which is consistent with the assumption you made.

Your instincts are a fine attempt, but notice that no x can satisfy both conditions. $$\log_{10}(x)=10 \quad and \quad\log_{10}(\log_{10}(x))=4$$

Your substitution would work, but it doesn't simplify the problem enough.

It's probably easier to use this substitution: $$Let\quad10^{10^a} = x$$ substitute: $$\log_{10}(10^{10^a})^{\log_{10}(\log_{10}(10^{10^a}))}= 10,000$$ simplify: $$(10^a)^a = 10^4$$ $$a = ±2\quad\Rightarrow\quad x=10^{10^{±2}}$$ so, $$x = 10^{100}\quad or \quad x = 10^{1/100}$$