SOLUTION: How would you solve: log(base 2x)+log(base 4x)+log(base 8x) = 1?

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Question 149609: How would you solve: log(base 2x)+log(base 4x)+log(base 8x) = 1?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%28x%29%29%2Blog%284%2C%28x%29%29%2Blog%288%2C%28x%29%29=1 Start with the given equation.



Use the Change of Base formula to rewrite each log. Remember, the Change of Base formula is: log%28b%2C%28x%29%29=log%2810%2C%28x%29%29%2Flog%2810%2C%28b%29%29


Rewrite 4 as 2%5E2. Rewrite 8 as 2%5E3.


Rewrite each log using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29


Now to make things simple, let z=log%2810%2C%282%29%29. So this means that the equation is now





Notice now that the LCD is 6z


Multiply both sides by the LCD 6z to clear the fractions.


6%2Alog%2810%2C%28x%29%29%2B3%2Alog%2810%2C%28x%29%29%2B2%2Alog%2810%2C%28x%29%29=6z Distribute and multiply.


log%2810%2C%28x%29%29%286%2B3%2B2%29=6z Factor out the GCF log%2810%2C%28x%29%29


log%2810%2C%28x%29%29%2811%29=6z Add


11%2Alog%2810%2C%28x%29%29=6z Rearrange the terms.


log%2810%2C%28x%29%29=%286z%29%2F%2811%29 Divide both sides by 11.


log%2810%2C%28x%29%29=%286%2Alog%2810%2C%282%29%29%29%2F%2811%29 Now replace "z" with log%2810%2C%282%29%29.


log%2810%2C%28x%29%29=%286%2F11%29%2Alog%2810%2C%282%29%29 Rearrange the terms.


log%2810%2C%28x%29%29%2Flog%2810%2C%282%29%29=6%2F11 Divide both sides by log%2810%2C%282%29%29.


log%282%2C%28x%29%29=6%2F11 Use the change of base formula to rewrite the left side.


Rewrite the equation using the property: log%28b%2C%28x%29%29=y ====> b%5Ey=x


So the answer is which approximates to x=1.45948

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log(base2) x+log(base4) x + log(base8) x = 1
----------------
[logx/log2] + [logx/2log2] + [logx/3log2] = 1
Multiply thru by 6log2 to get:
6logx + 3logx + 2logx = 6log2
11logx = 6log2
[logx/log2] = 6/11
log(base2) x = 6/11
x = 2^(6/11)
x = 1.45948...
-----------------
Cheers,
Stan H.