3 = log8 + 3logx need help please and thankyou

Question

anonymous · Answer

start by using $n\log(x)=\log(x^n)$ to rewrite it as 
$$3=\log(8)+\log(x^3)$$

anonymous · Answer

then use $\log(A)+\log(B)=\log(AB)$ to rewrite it as 
$$3=\log(8x^3)$$

mathslover · Answer

$\log a + \log b = \log (ab) $ - Next step

callisto · Answer

log 8 = log (2^3) = 3 log 2

3 = log8 + 3logx
3 = 3 log 2 + 3 log x
1 = log 2 + log x

mathslover · Answer

here, a = 8 and b = x^3 ...

anonymous · Answer

then rewrite in equivalent exponential form
assuming that the log is log base ten, exponential form is $$10^3=8x^3$$

anonymous · Answer

then your puttin 3=x+3

mathslover · Answer

Next step : 

$3 = \log (2x)^3 $ 

$3 = 3 \log (2x) $

$1 = log (2x)$

anonymous · Answer

solve for $x$ by dividing by $8$ and then taking the cubed root of both sides

mathslover · Answer

That is : $1 = \log (2x) $ 

If base is 10, then 2x = 10

as : $log_ a a = 1$

anonymous · Answer

x=5

mathslover · Answer

solve it for x further. 

$\bf{2x  = 10}$

anonymous · Answer

$$10^3=8x^3$$
$$\frac{10^3}{8}=x^3$$
$$\frac{10}{2}=x$$$$5=x$$

mathslover · Answer

Simply use : $\log _ a a = 1$ , if base is 10, then 2x has to be 10 also. That's it...