Stacking Logs Side By Side

Set $\log_3 x=a$ , $\log_4 x=b$ , $\log_5 x=c$ . Then we have: $abc=ab+bc+ca$ Suppose $a,b,c\neq 0$ . Then dividing by $abc$ gives: $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$ Note that: $\dfrac{1}{a}=\dfrac{1}{\log_3 x}=\log_x 3$ . Similarly, $\dfrac{1}{b}=\log_x 4$ and $\dfrac{1}{c}=\log_x 5$ . So we have: $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\log_x 3+\log_x 4+\log_x 5=\log_x 60=1 ~~~\Longrightarrow ~ x=60$ When one of $a,b,c$ is $0$ , suppose $a=0$ , then $\log_3 x=0$ which implies $x=1$ . So the answer is: $60+1=61.$

Change the basis, where

$\log_{3} x =\frac{\ln x}{\ln 3}$ , $\log_{4} x =\frac{\ln x}{\ln 4}$ , $\log_{5} x =\frac{\ln x}{\ln 5}$

then,

$\log_{3} x \log_{4} x \log_{5} x = \log_{3} x \log_{4} x + \log_{4} x \log_{5} x + \log_{5} x \log_{3} x$

will become

$\frac{\ln x}{\ln 3} \frac{\ln x}{\ln 4} \frac{\ln x}{\ln 5} = \frac{\ln x}{\ln 3} \frac{\ln x}{\ln 4} + \frac{\ln x}{\ln 4} \frac{\ln x}{\ln 5} + \frac{\ln x}{\ln 5} \frac{\ln x}{\ln 3}$

$\frac{[\ln x]^3}{\ln 3 \ln 4 \ln 5} = \frac{[\ln x]^2}{\ln 3 \ln 4} + \frac{[\ln x]^2}{\ln 4 \ln 5} + \frac{[\ln x]^2}{\ln 5 \ln 3}$

$[\ln x]^3 = \ln 5 [\ln x]^2 + \ln 3 [\ln x]^2 + \ln 4 [\ln x]^2$

$[\ln x]^3 - \ln 5 [\ln x]^2 - \ln 3 [\ln x]^2- \ln 4 [\ln x]^2 = 0$

$[\ln x]^2 [ \ln x - \ln 5 - \ln 3 - \ln 4 ] = 0$

We have two factors,

$[\ln x]^2 = 0$

where we will get $x = 1$

and

$\ln x - \ln 5 - \ln 3 - \ln 4 = 0$

$\ln x = \ln 5 + \ln 3 + \ln 4$

$\ln x = \ln 60$

$x = 60$

therefore we get 60 + 1 = 61

For this problem, I used change of base and factoring on the original problem to get

((Log x)^3)/(log 3 log 4 log 5) = [log x]^2 (log 3 log 4 + log 4 log 5 + log 5*log 3)

Clearing the fraction with lcd

(Log x)^3 = [log x]^2 (log 5 + log 4 + log 3)

Using properties of logs

(Log x)^3 = [log x]^2 (log 60)

Moving all terms to other side and factoring we get

(Log x)^2 (log x - log 60) = 0

Hence log x = 0, x = 1, log x = log 60, x = 60

And so 1 + 60 = 61, our solution