Find $x$, where $3^{\log_6x} + 4^{\log_6x} + 5^{\log_6x} = x$ Find $x$, where $3^{\log_6x} + 4^{\log_6x} + 5^{\log_6x} = x$. As solutions, my workbook gives the following choices:
a) $x\in (0,60)$, b)$x \in(60,100)$, c) $x \in (100, 200)$, d) $x \in (200,300)$, e) $x \in (300, \infty)$
What I tried so far is to switch the arguments of the logarithms with the base of the exponentials and got:
$$x^{\log_63} + x^{\log_64} + x^{\log_65}= x$$
At this point, being stuck but given the a-e choices, I tried to guess the answer, but I was totally off. Without a calculator which I assume could help with finding the values of the logarithms, how would I go about finding $x$ here? Is there a way to approach such problems to find exactly the solution or at least a very narrow interval? Could analysis be used to speed up getting an approximate value of $x$?
 A: Given that you are offered intervals as choices for where the value of $ \ x \ $ lies, you are likely correct that you only need an estimate.  So it is reasonable to apply the "change-of-base formula" to produce, as you did,
$$ 3^{\log_6 x} \ + \ 4^{\log_6 x} \ +  \ 5^{\log_6 x} \ \ = \ \ 3^{\log_6 (3)·\log_3 x} \ + \ 4^{\log_6 (4)·\log_4 x} \ +  \ 5^{\log_6 (5)·\log_5 x} $$ $$ = \ \ x^{\log_6 (3)} \ + \ x^{\log_6 (4)} \ +  \ x^{\log_6 (5)} = \ \  x  \ \ . $$
We might make some (crude) estimates of these logarithms:
$$ 6^5 \ \ = \ \ 7776 \ \ \ , \ \ \ 6^6 \ \ = \ \ 46,656 \ \ \ , \ \ \ 6^7 \ \ = \ \ 279,936 \ \ ; $$
$$ 3^8 \ \ = \ \ 6561 \ \ \ , \ \ \ 4^8 \ \ = \ \ 65,536 \ \ \ , \ \ \ 5^8 \ \ = \ \ 390,625   $$
$$ \Rightarrow \ \ \log_6(3) \ \ \sim \ \ \frac58 \ \ \ , \ \ \ \log_6(4) \ \ \sim \ \ \frac68 \ \ \ , \ \ \ \log_6(5) \ \ \sim \ \ \frac78 \ \ \ ,    $$
producing an estimate for our "equation"
$$  x^{5/8} \ + \ x^{6/8} \ +  \ x^{7/8} \ \ = \ \ x^{5/8}   · ( \ 1 + \ x^{1/8} \ +  \ x^{2/8} \ ) \ \ \sim \ \  x  \ \  \Rightarrow \ \ 1 \ + \ x^{1/8} \ +  \ x^{2/8} \   \ \ \sim \ \  x^{3/8} \ \ . $$
If we now take $  \ u \ = \ x^{1/8} \ \ , \ \ $ we see that $ \ u^3 \ \sim \ u^2 \ + \ u \ + \ 1 \ $ can be "satisfied" by $ \ u \ \sim \ 2 \ \ , \ $ giving us $ \ x \ \sim \ 2^8 \ = \ 256 \ \ , \ $ which falls into the interval $ \ (200 \ , \ 300) \ \ $ [choice $ \ \mathbf{(d)} \ ] \ . $  (More precise values of the logarithms are $ \ 0.613 \ , \ 0.774 \ , \ $ and $ \ 0.898 \ \ ; \ $ more precise rational approximations, however, make the estimation harder to work with for a small improvement in the result for $ \ x \ \ . \ ) $
This method is within reach without a calculator if one is reasonably comfortable with "arithmetic by hand" (not that I'd want to have to do that for an exam) or has some "number facts" at hand (one could also apply $$ \log_{10}(3) \ \sim \ 0.5 \ \ , \ \ \log_{10}(4) \ \sim \ 0.6 \ \ , \ \ \log_{10}(5) \ \sim \ 0.7 \ \ , \ \ \log_{10}(6) \ \sim \ 0.8 \ \ . \ \ ) $$
It is of course very nice, if not immediately obvious, that the given equation has an integer solution of $ \ x \ = \ 216 \ \ . $ [Kudos to
Adam Rubinson for spotting that!]
