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

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) (Show Source): You can put this solution on YOUR website!
Start with the given equation.

Use the Change of Base formula to rewrite each log. Remember, the Change of Base formula is:

Rewrite as . Rewrite as .

Rewrite each log using the identity

Now to make things simple, let . So this means that the equation is now

Notice now that the LCD is 6z

Multiply both sides by the LCD to clear the fractions.

Distribute and multiply.

Factor out the GCF

Add

Rearrange the terms.

Divide both sides by 11.

Now replace "z" with .

Rearrange the terms.

Divide both sides by .

Use the change of base formula to rewrite the left side.

Rewrite the equation using the property: ====>

So the answer is which approximates to
Answer by stanbon(75887) (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.
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