SOLUTION: I'm not really sure how to approach these questions:
log(base 3)(2x^2 + 4) ......Solving in terms of common log
...And...
Log(2x - 1) - log(4x + 3) = 1 ........Solving f
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-> SOLUTION: I'm not really sure how to approach these questions:
log(base 3)(2x^2 + 4) ......Solving in terms of common log
...And...
Log(2x - 1) - log(4x + 3) = 1 ........Solving f
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Question 634689: I'm not really sure how to approach these questions:
log(base 3)(2x^2 + 4) ......Solving in terms of common log
...And...
Log(2x - 1) - log(4x + 3) = 1 ........Solving for x.
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I have two textbooks and neither of them show examples of how to work similar problems, please help. Found 2 solutions by Alan3354, lwsshak3:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! log(base 3)(2x^2 + 4) ......Solving in terms of common log
Not an equation, nothing to solve.
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...And...
Log(2x - 1) - log(4x + 3) = 1
Subtracting logs --> division
log((2x-1)/(4x+3)) = 1
(2x-1)/(4x+3) = 10
etc.
You can put this solution on YOUR website! I'm not really sure how to approach these questions:
log(base 3)(2x^2 + 4) ......Solving in terms of common log
...And...
Log(2x - 1) - log(4x + 3) = 1 ........Solving for x.
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When working with log problems it is very important to identify the following:
base, log of the number, and the number itself. Then there are the rules for multiplication, division and exponents. Logs don't work with addition and subtraction.
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For your first question: There is a formula for changing the base and I'm sure it is somewhere in your text. This is how you change bases:
Take the log of the number with the new base and divide by the log of the original base with the new base. In your example:
log3(2x^2+4)=log(2x^2+4)/log3
you can easily do this on the calculator changing to common or natural(ln) logs
..
Your 2nd question:
Log(2x - 1) - log(4x + 3) = 1
place under single log
log[(2x-1)/(4x+3)]=1
subtracting logs means numbers are dividing, adding logs means multiplication.
convert to exponential form: base(10) raised to log of the number(1)=(2x-1)/(4x+3)
10^1=(2x-1)/(4x+3)=10
10(4x+3)=2x-1
40x+30=2x-1
38x=-31
x=-31/38
x≈-.816
no solution: (2x-1)>0
note: logs>0
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