Solve the following equation : $\log_2(x)*\log_4(x)*\log_8(x)=4.5$ I have the following equation :
$$\log_2(x)*\log_4(x)*\log_8(x)=4.5$$
Usually, I do post what I made to do, but in this case a friend of mine tackle me with this question after I didn't mess with these kind of equations more than two years so I forgot how to approach these kind of equations, I'll be glad if someone could solve it so I'll learn from the solution.
Any help will be appreciated, thanks.
 A: Use the fact that $\log_a(b)=\frac{\log_c(b)}{\log_c(a)}$. 
This gives $$\log_2(x)\log_4(x)\log_8(x) = \log_2(x)\frac{\log_2(x)}{\log_2(4)}\frac{\log_2(x)}{\log_2(8)}=\frac16(\log_2(x))^3$$
So $\frac16(\log_2(x))^3=4.5$, hence $(\log_2(x))^3=27$, hence $\log_2(x)=3$, and thus $x=8$. It can be seen that $x=8$ indeed satisfies the given equation. 
A: Here's one approach:
Note: $\log_a a = 1$, $\log_b a = \frac{\log_n a}{\log_n b}$
$$ \log_2 x\times \log_4 x \times \log_8 x = 4.5 = \frac{9}{2}$$
change all logs to base 2:
$$\log_2 x \times \frac{\log_2 x}{\log_2 4}\times \frac{\log_2 x}{\log_2 8} = \frac{9}{2} $$
$$ \log_2 x \times \frac{1}{2}\log_2 x \times \frac{1}{3}\log_2 x = \frac{9}{2}$$
$$(\log_2 x )^3 = 27$$
I trust you can finish it from here.
A: it is $$\frac{\ln(x)}{\ln(2)}\frac{\ln(x)}{2\cdot\ln(2)}\frac{\ln(x)}{3\ln(2)}=\frac{9}{2}$$ and from here we get
$$\frac{\ln(x)^3}{6\cdot (\ln(2))^3}=\frac{9}{2}$$ calculating this we obtain $$\ln(x)^3=27\cdot \ln(2)^3$$ can you proceed?
A: Hint: Use the change of base formula to see
$$
\log_4 x = {\log_2 x\over \log_2 4} = {\log_2 x \over 2}
$$
Similarly
$$
\log_8 x = {\log_2 x\over 3}
$$
A: $$\log_2(x)∗\log_4(x)∗\log_8(x)=4.5$$
Using the log change-of-base rule $\log_c(x) = \frac{\log_b(x)}{\log_b(c)}$ and setting $b = 3$, we get
$$\frac{\log_3^3(x)}{\log_3(2)*\log_3(4)*\log_3(8)} = 4.5$$
Then, using the log power rule $\log_c(a^b) = b*\log_c(a)$, we get
$$\frac{\log_3^3(x)}{\log_3^3(2)} = 27$$
proceeding to be
$$\frac{\log_3(x)}{\log_3(2)} = 3$$
But then, you just multiply by the denominator on each side to get
$$\log_3(x) = \log_3(8)$$
A: This equation is equivalent to :
$$\log_{2}(x)\frac{\log_{2}(x)}{2}\frac{\log_{2}(x)}{3} = 4.5$$
Because of : $$\log_{m^{n}}(x)= \frac{\log_{m}(x)}{n}$$
A: Raise all the logs to the same base, by raising the argument of each log by the necessary power.
$\log_2{x}\times\log_4{x}\times\log_8{x}=4.5$
$\log_8{x^3}\times\log_8{x^{3/2}}\times\log_8{x}=4.5$
$3\times(3/2)\times\log_8{x}\times\log_8{x}\times\log_8{x}=4.5$
$(log_8{x})^3=1$
$x=8$
