A question on Logarithms

Question

Q:

Given that $\log_3(x) = a$ solve for $x$,

$\log_3(9x) + \log_3(\frac{x^3}{81}) = 3$

\I make progress by writing $\log_3(9x) = 3^{2+a}$ and $\log_3(\frac{x^3}{81}) = 5a - 4$.

However, I can't finish it off.

Thanks

Answer 1

Logaritm rules gives that $$ \log_3(9x) = \log_3(9) + \log_3(x) = 2 + a $$ and that $$ \log_3\left(\frac{x^3}{81}\right) = \log_3(x^3) - \log_3(81) = 3\log_3(x) - 4 = 3a - 4 $$ Insert this into your equation, solve for $a$, then use $\log_3(x) = a$ to solve for $x$.

Answer 2

The easiest way to start this problem is to simplify the equation using properties of logarithms (I think you may have applied them incorrectly). The term $\log_3(9x)$ equals $\log_3(9) + \log_3(x)=2 + a$, and the term $\log_3(x^3/81)$ equals $3\log_3(x) - \log_3(81)=3a-4$. Now solve for $a$, and from there solve for $x$.

Answer 3

First of all, use the logarithm rule

lg(u) + lg(v) = lg(uv)

So , you get

$$log_3(\frac{x^4}{9}) = 3$$

Take 3^ on both sides to get

$$\frac{x^4}{9} = 3^3 = 27$$

So, finally , we get

$$x = 3^\frac{5}{4}$$

because x has to be positive.