Logs and Logs!

Algebra Level 2

If a , b , c > 0 and x > 1 a,b,c>0 \text{ and } x>1 , let

{ log a b x = 9 log b c x = 18 log a b c x = 8 \begin{cases} \log_{ab}x=9 \\ \log_{bc}x=18 \\ \log_{abc}x=8 \end{cases}

Then what is the value of log a c x = ? \log_{ac}x=?

12 36 18 9 24

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3 solutions

Pi Han Goh
May 13, 2019

Let y = log a c x y = \log_{ac} x .

Using the logarithmic property, log p q = 1 log q p \log_p q = \frac1{\log_q p} , we can rewrite all the given equations as { log x a b = 1 9 log x b c = 1 18 log x a c = 1 y log x a b c = 1 8 . \begin{cases} \log_ x ab = \frac19 \\ \log_x bc = \frac1{18} \\ \log_x ac = \frac1y \\ \log_x abc = \frac18. \end{cases}

And because log p q + log p r = log p q r \log_p q + \log_p r = \log_p qr , the sum of the first 3 given equations is equal to twice the remaining given equation, log x a b + log x b c + log x a c = 2 log x a b c 1 9 + 1 18 + 1 y = 2 8 y = 12 . \log_ x ab + \log_x bc + \log_x ac = 2 \log_x abc \quad \Leftrightarrow \quad \frac19 + \frac1{18} + \frac1y = \frac28 \quad \Leftrightarrow \quad y = \boxed{12} .

Thank you.

Hana Wehbi - 2 years ago
Chew-Seong Cheong
May 13, 2019

{ log a b x = 9 x = ( a b ) 9 a b = x 1 9 . . . ( 1 ) log b c x = 18 x = ( b c ) 18 b c = x 1 18 . . . ( 2 ) log a b c x = 8 x = ( a b c ) 8 a b c = x 1 8 . . . ( 3 ) \begin{cases} \log_{ab} x = 9 & \implies x = (ab)^9 & \implies ab = x^{\frac 19} & ...(1) \\ \log_{bc} x = 18 & \implies x = (bc)^{18} & \implies bc = x^{\frac 1{18}} & ...(2) \\ \log_{abc} x = 8 & \implies x = (abc)^8 & \implies abc = x^{\frac 18} & ...(3) \end{cases}

{ ( 3 ) ( 2 ) : a = x 5 72 ( 3 ) ( 1 ) : c = x 1 72 a c = x 1 12 x = ( a c ) 12 log a c x = 12 \implies \begin{cases} \dfrac {(3)}{(2)}: & a = x^\frac 5{72} \\ \dfrac {(3)}{(1)}: & c = x^\frac 1{72} \end{cases} \implies ac = x^\frac 1{12} \implies x = (ac)^{12} \implies \log_{ac} x = \boxed{12}

Thank you Sir.

Hana Wehbi - 2 years ago

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You are welcome.

Chew-Seong Cheong - 2 years ago
Katyusha Le
May 13, 2019

I hope this will help.

Thanks for sharing your solution.

Hana Wehbi - 2 years, 1 month ago

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