# Logs...

• August 3rd 2006, 04:17 PM
juangohan9
Logs...
You guys were pretty good last time, so I wanted to see what your thoughts were on these logs...

I have to find x for this equation...

logbase9*(x-8) + logbase9 x = 1

do I have 2x? And then, how do I get rid of logbase9?

This one also ****** me off, 4^4 logbase4^4...

so it's 4, to the 4logbase4, to another power of 4... errr...

last is...

(6/7)^x+1 = 343/216

How do I get rid of that (6/7) ?

Thanks a lot, again.
• August 3rd 2006, 04:42 PM
ThePerfectHacker
Quote:

Originally Posted by juangohan9
You guys were pretty good last time, so I wanted to see what your thoughts were on these logs..

I have to find x for this equation...

logbase9*(x-8) + logbase9 x = 1

Use the rule,
$\log_b(xy)=\log_b x+\log_b y$
Thus,
$\log_9 (x-8)+\log_9 x$
Becomes,
$\log_9 [x(x-8)]=1$
By the defintion of logarithms,
$9^1=x(x-8)$
Thus,
$x^2-8x=9$
Thus,
$x^2-8x-9=0$
Thus,
$(x-9)(x+1)=0$
Thus,
$x=9,-1$
But,
$x\not =-1$ because logarithms are not definied for negative numbers. Thus, $x=9$

Quote:

4^4 logbase4^4...

so it's 4, to the 4logbase4, to another power of 4... errr...
Easy $4^4=256$
So you have,
$256\log_4 4^4$
Now what is? $\log 4^4$
Use the rule,
$\log_b b^n=n$
Thus,
$256 \cdot 4= 1024$

Quote:

(6/7)^x+1 = 343/216

How do I get rid of that (6/7) ?

Thanks a lot, again.
I think you mean,
$(6/7)^x=343/216$, right?
• August 3rd 2006, 04:49 PM
topsquark
Quote:

Originally Posted by juangohan9
You guys were pretty good last time, so I wanted to see what your thoughts were on these logs...

I have to find x for this equation...

logbase9*(x-8) + logbase9 x = 1

do I have 2x? And then, how do I get rid of logbase9?

This one also ****** me off, 4^4 logbase4^4...

so it's 4, to the 4logbase4, to another power of 4... errr...

last is...

(6/7)^x+1 = 343/216

How do I get rid of that (6/7) ?

Thanks a lot, again.

$log_9(x-8) + log_9(x) = 1$

$log_9(x^2-8x) = 1$

Thus
$x^2 - 8x = 9^1 = 9$
$x^2 - 8x - 9 = (x-9)(x+1) = 0$

Thus x = 9, -1. However note that x = -1 gives an impossibility: $log_9(-1-8) = log_9(-9)$ in the original equation is undefined. Thus x = 9 is the only solution.

I'm not quite sure what you mean by "4^4 logbase4^4." I can't make your description of it make sense. Is it a typo or am I seeing something wrong here?

I presume the last one is supposed to be:
$(6/7)^{x+1} = 343/216$

Note that 6^3 = 216 and 7^3 = 343. Thus
$(6/7)^{x+1} = (7/6)^3 = (6/7)^{-3}$

Thus
$x + 1 = -3$

and x = -4.

-Dan
• August 3rd 2006, 04:50 PM
topsquark
Hmmm...TPH and I seem to be busy with all the same posts today...

-Dan