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solve for x : log base 9 X^2=9

Guest Aug 6, 2015

Best Answer 

 #2
avatar+20711 
+10

 $$\small$\text{
Solve for x : $\log_9{(x^2)}=9$
}}$$

 

$$\small{\text{Formula: $\boxed{~ \log_b{(b^a)}=a~}$}}\\\\
\small{\text{
We have $b=9$, and $a = 9$ so $\log_9{(9^9)}=9$, but $9^9$ is $x^2$
}}\\\\
\small{\text{$
\begin{array}{rcl}
x^2 &=& 9^9\\
x &=& 9^{\frac{9}{2}}\\
x &=& \left(\sqrt{9} \right)^9\\
x &=& \left(\sqrt{3^2} \right)^9\\
x &=& 3^9\\
\mathbf{x} & \mathbf{=} & \mathbf{19683}\\
\\
\hline
\\
\end{array}
$}}\\$$

 

heureka  Aug 6, 2015
 #1
avatar+94202 
+10

$$\\log_9x^2=9\\\\
2log_9x=9\\\\
log_9x=9/2\\\\
9^{log_9x}=9^{9/2}\\\\
x=9^{9/2}\\\\
x=3^9\\\\
x=19683$$

 

 

check:

$${{log}}_{{\mathtt{9}}}{\left({{\mathtt{19\,683}}}^{{\mathtt{2}}}\right)} = {\mathtt{9}}$$

Melody  Aug 6, 2015
 #2
avatar+20711 
+10
Best Answer

 $$\small$\text{
Solve for x : $\log_9{(x^2)}=9$
}}$$

 

$$\small{\text{Formula: $\boxed{~ \log_b{(b^a)}=a~}$}}\\\\
\small{\text{
We have $b=9$, and $a = 9$ so $\log_9{(9^9)}=9$, but $9^9$ is $x^2$
}}\\\\
\small{\text{$
\begin{array}{rcl}
x^2 &=& 9^9\\
x &=& 9^{\frac{9}{2}}\\
x &=& \left(\sqrt{9} \right)^9\\
x &=& \left(\sqrt{3^2} \right)^9\\
x &=& 3^9\\
\mathbf{x} & \mathbf{=} & \mathbf{19683}\\
\\
\hline
\\
\end{array}
$}}\\$$

 

heureka  Aug 6, 2015

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