My Math Forum Bogged in logs

 Linear Algebra Linear Algebra Math Forum

 January 8th, 2017, 12:00 PM #1 Member   Joined: Nov 2016 From: Ireland Posts: 79 Thanks: 2 Bogged in logs I'm fairly familiar with logs - or so I thought. There's something tripping me up here though. Perhaps its the algebra. Something... anyways $\displaystyle log_4 x - log_4 (x-1) = 2$ I've got it to $\displaystyle log_4 x/x-1 = 2$ Think I'm messing something up after that. I've tried transferring the power across $\displaystyle 4^2$ and working at it like that. Keep coming back with 0.6...... or even -0.6.... when I substitute the answer back in. I'm a lil burned out at this stage think I'm messing up the algebra.
 January 8th, 2017, 12:08 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 First, note that $x > 1$ $\log_4\left(\dfrac{x}{x-1}\right)=2$ $4^2 = \dfrac{x}{x-1}$ $16(x-1)=x$ $16x-16=x$ $15x=16$ $x=\dfrac{16}{15} >1$ Thanks from Kevineamon
 January 8th, 2017, 12:19 PM #3 Member   Joined: Nov 2016 From: Ireland Posts: 79 Thanks: 2 Dang couldn't see the wood for the trees. Why it always look so simple when you guys do it? It's like watching pro golf. Anybody fancy lending me their brain for a day? Give you a good price for it. Sigh... thanks again Skeeter

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