solve this logarithm

esatpllana · 2008-03-14 23:17:56

thanks

luca-deltodesco · 2008-03-15 01:10:22

perhaps someone else can show how to get x analytically here i can get as far as:

but i can't figure out how to show that the left side is equal to x, without already assuming that it IS equal to 9. or is that ok? im confused and drunk leave me be.

Last edited by luca-deltodesco (2008-03-15 01:18:07)

luca-deltodesco · 2008-03-15 02:12:27

Last edited by luca-deltodesco (2008-03-15 02:14:26)

luca-deltodesco · 2008-03-15 02:26:06

also, why do people find it necessary to come up with such silly equations to give to students to solve?

esatpllana · 2008-03-15 02:35:26

comment this photo

Last edited by esatpllana (2008-03-15 02:41:54)

John E. Franklin · 2008-03-15 05:08:02

The 9 is

John E. Franklin · 2008-03-15 05:16:34

I will double check
luca's work with a
funny notation for
the fun of it.

Legend:
  L3G = log base 3
  L9G = log base 9
  xUP2 = x times x = x multiplied by x = x^2 = x to the power of 2 = (x)(x)
  xDOWN2 = square root of x = vx = xUP0.5
  xOVER3 = x/3 = x divided by 3
  xMULT3 = 3x = 3 times x = x + x + x = 3 + 3 + 3, x number of times.

Here I go:
L3G{1 OVER [(L3Gx)DOWN2]} = L9G L9G(x OVER 3)

Change L3G to L9G by squaring stuff inside.

L9G(1 OVER L3Gx) = sameAsAbove

Rid outer L9G on both sides of equation keeping in mind
that answers may be invalid or undefined due to logs of
negatives being bad.

1 OVER L3Gx = L9G(x OVER 3)

Get both sides into base 9 logs.

1 OVER (L9G xUP2) = sameAsAbove

Multiply both sides by denominator of left side.

1 = (L9G xUP2) MULT L9G(x OVER 3)

The log of something squared is double the log of something.

1 = 2 MULT L9Gx MULT L9G(x OVER 3)

The L9G of something over 3 is one-half less than the L9G of
said something because 9 UP (-1 OVER 2) = 1 OVER 3.

1 = 2 MULT L9Gx MULT {-[1 OVER 2] + L9Gx}

Use substitution for a while: Let H = L9Gx

1 = 2 MULT H MULT {-[1 OVER 2] + H}

Multiply together (expand right side) and get 0 on the left side.

1 = 2 MULT (H UP 2) - H

0 = 2 MULT (H UP 2) - H - 1

use quadratic formula (or could factor, but I'm bad at that)
a = 2,  b = -1, c = -1

"OR" means 2 answers

H = [-b plusORminus ( bUP2 - 4 MULT a MULT c) DOWN 2] OVER (2 MULT a)

H = [+1 plusORminus (-1UP2 - 4 MULT 2 MULT{-1})DOWN 2] OVER (2 MULT 2)

H = [ 1 plusORminus (  1   - 4 MULT 2 MULT(-1))DOWN 2 ] OVER 4

H = [ 1 plusORminus (  1   + 4 MULT 2 MULT(+1))DOWN 2 ] OVER 4

H = [ 1 plusORminus (  1   + 4 MULT 2 MULT(+1))DOWN 2 ] OVER 4

H = [ 1 plusORminus (  1   + 8 )DOWN 2 ] OVER 4

H = [ 1 plusORminus (  9 )DOWN 2 ] OVER 4

H = [ 1 plusORminus 3 ] OVER 4

H = [ 4 OR -2 ] OVER 4

H = (4 OVER 4) OR H = -2 OVER 4

H = 1  OR  H = -1 OVER 2

Revert H back to L9Gx

L9Gx = 1  OR  L9Gx = -1 OVER 2

9 UP 1 = x   OR  9 UP (-1 OVER 2) = x

x = 9  or x = 3 UP -1

x = 9 or x = 1 OVER 3

Same answer as luca-delt.

But throw away the 1 OVER 3 because it is invalid with original equation
due to logs of negatives coming in when we removed logs on both sides
of equation, a risky maneuver.

So x = 9.

Last edited by John E. Franklin (2008-03-15 07:01:31)

Kurre · 2008-03-16 05:53:02

heres a solution:

raise both sides with 9:

9=3^2 so...

we have the identity which gives us:

substituting and multiplication with 2y and rearranging gives the equation:

which has the roots 2 and -1.
y=2 gives x=9 which works with the original equation, but is not defined for log3x=-1 (or if you want complex numbers, the logarithm of an imaginary number does not work)
hence, x=9 is the only solution

Math Is Fun Forum

#1 2008-03-14 23:17:56

solve this logarithm

#2 2008-03-15 01:10:22

Re: solve this logarithm

#3 2008-03-15 02:12:27

Re: solve this logarithm

#4 2008-03-15 02:26:06

Re: solve this logarithm

#5 2008-03-15 02:35:26

Re: solve this logarithm

#6 2008-03-15 05:08:02

Re: solve this logarithm

#7 2008-03-15 05:16:34

Re: solve this logarithm

#8 2008-03-16 05:53:02

Re: solve this logarithm

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