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October 23, 2014

October 23, 2014

Posted by **Lea** on Monday, January 30, 2012 at 11:39pm.

2. Solve Log[8](log[8]x)=0

- Precalculus -
**Reiny**, Tuesday, January 31, 2012 at 12:43amlet 5^x = y , then we have

y^2 - 4y - 12 = 0

(y-6)(x+2) = 0

y = 6 or y = -2

5^x = 6

xlog5 = log6

x = log6/log5 = appr. 1.1133

or

5^x = -2

xlog5 = log (-2), but we cannot take logs of negatives, so no solution for that case

x = log6/log5 or 1.1133

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