Solving for $x$

Algebra Level 2

Find the value of $x$ that satisfies $\log_{3}\left(\log_{9} x\right) = \log_{9}\left(\log_{3} x\right)$ .

3 solutions

David Vreken
Mar 24, 2019

Let $x = 9^n$ . Then

$\log_3 (\log_9 x) = \log_9 (\log_3 x)$

$\log_3 (\log_9 9^n) = \log_9 (\log_3 9^n)$

$\log_3 (\log_9 9^n) = \log_9 (\log_3 3^{2n})$

$\log_3 n = \log_9 2n$

$9^{\log_3 n} = 2n$

$3^{2\log_3 n} = 2n$

$3^{\log_3 n^2} = 2n$

$n^2 = 2n$

Since $n \leq 0$ gives an extraneous solution for $x$ , $n = 2$ , which means $x = 9^2 = \boxed{81}$ .

Nice solution. Thank you.

Hana Wehbi - 2 years, 2 months ago

Joshua Lowrance
Mar 23, 2019

$\log_{3}{(\log_{9}{x})}=\log_{9}{(\log_{3}{x})}=$

$9^{\log_{3}{(\log_{9}{x})}}=9^{\log_{9}{(\log_{3}{x})}}$

$3^{2\log_{3}{(\log_{9}{x})}}=\log_{3}{x}$

$3^{\log_{3}{(\log_{9}{x})^2}}=\log_{3}{x}$

$(\log_{9}{x})^2=\log_{3}{x}$

$( \frac{ \log_{3}{x}}{ \log_{3}{9}} )^2 = \log_{3}{x}$

$\frac{ ( \log_{3}{x})^2}{2^2} = \log_{3}{x}$

$\frac{1}{4}=\frac{1}{\log_{3}{x}}$

$x=81$

Nice solution. Thank you.

Hana Wehbi - 2 years, 2 months ago

Avishai Feuer
Mar 24, 2019

But if x=1 u will get
-∞=-∞

-∞ is not a number, so you can't say -∞=-∞ is true because you can't equate two non-numbers together.

Pi Han Goh - 2 years, 2 months ago