Solve Logarithmic Equation: 4(x+1)= 8(x-1)

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The discussion revolves around solving the logarithmic equation 4(x+1) = 8(x-1). The initial approach involved taking logarithms, leading to an expression for x that was calculated as approximately 5. Participants noted that using base 2 logarithms simplifies the problem significantly, transforming it into a more manageable equation. Alternative methods were also discussed, including manipulating the equation directly without logarithms by equating exponents after rewriting the bases. The conversation emphasizes the importance of recognizing properties of logarithms and bases to solve such equations effectively.
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Just wondering if someone can tell me if I got this correct. It was a problem on my midterm a month ago and I am going over the probs I got wrong, and I don't know the correct answers.

Homework Statement



4(x+1)= 8(x-1)

Homework Equations





The Attempt at a Solution


log 4(x+1) = log 8(x-1)

(x+1) log 4 = (x-1) log 8

x log 4 + log 4 = x log 8 - log 8 (sub/add to get x on one side)

log 8 + log 4 = x log 8 - x log 4

log 8 + log 4 = x(log 8 - log 4) (divide to get x)

log 8 + log 4 = X
---------------
log 8 - log 4

x= 5
 
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I guess that works, though you didn't really say how you changed your final log expression into 5. But you can make your life a bit easier by picking a convenient base for the logarithms. In this case I'd pick log_2 (log base 2) and use that. So your second line just turns into 2(x+1)=3(x-1).
 
^
well, that would make it easier, now wouldn't it? prob why i got it wrong on the exam. i totally forgot about that.

To get 5, I divided
(log 8 + log 4)/(log 8 - log 4)
which is
1.505149978/.3010299957 = 4.999999998

thank you though for reminding me about the bases. I couldn't do it the way I did it if it was on the non calculator portion of the exam, so I am going to have to remember that. I have another log question, but its a diff problem, so ill make another post.
 
You could also do it this way. (log(8)+log(4))=log(8*4)=log(32)=log(2^5)=5*log(2) and (log(8)-log(4))=log(8/4)=log(2). So you can do it without a calculator, though it pays to get rid of the logs early.
 
ah^
that is like the other question i posted. I forgot on the exam that addition is multiplication. Well, not on the expanding log questions but on solve for x questions.
 
Use of logarithms is not necessary for that example.
(2^2)^(x+1) Left Hand Side.
(2^3)^(x-1) Right Hand Side.

Skipping one step,
2^(2x+2) = 2^(3x-3)
Left and Right sides having the same base, the exponents are equal.
 
There's more than one way to skin a cat.
 
Yep, here's yet another one;
4^(x+1) = 8^(x-1)
16 * 4^(x - 1) = 8^(x - 1)
16 = (8/4)^(x - 1) = 2^(x - 1)
x - 1 = 4

Of course, most methods mentioned are not general, because they rely on the convenient fact that 4= 2^2 and 8 = 2^3.
 
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