logarithms

davehogan

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I have to solve the following equation using logs: "ex = pi". I then get "x = log pi/In e" but "In e = 1" so I end up with "x = log pi" = 0.497". The answer given is 1.145. Where am I going wrong?
 
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Subhotosh Khan

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I have to solve the following equation using logs: "ex = pi". I then get "x = log pi/In e" but "In e = 1" so I end up with "x = log pi" = 0.497". The answer given is 1.145. Where am I going wrong?
x = Ln(π) / Ln(e)
 

davehogan

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x = Ln(π) / Ln(e)

I see. Does that mean that I should not use "log pi" in this case, I should be using "Ln pi"? You must be right because "Ln pi" comes up with right answer.
 
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JeffM

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I see. Does that mean that I should not use "log pi" in this case, I should be using "Ln pi"? You must be right because "Ln pi" comes up with right answer.
Your error goes all the way back to the beginning.

\(\displaystyle x = y \implies log_z(x) = log_z(y).\)

Remember about changing BOTH sides of the equation equally. When you take logs of both sides of the equation, you must take logs using equal bases.

What you did was

\(\displaystyle e^x = \pi \implies log_e(e^x) = log_{10}( \pi ) \implies xlog_e(e) \approx 0.497 \implies x * 1 \approx 0.497 \implies x \approx 0.497.\)

That is WRONG because you took logarithms to different bases. The simple correct way is

\(\displaystyle e^x = \pi \implies log_e(e^x) = log_e( \pi ) \implies xlog_e(e) \approx 1.145\implies x * 1 \approx 1.145 \implies x \approx 1.145.\)

Another correct but somewhat more cumbersome way is

\(\displaystyle e^x = \pi \implies log_{10}(e^x) = log_{10}( \pi ) \implies xlog_{10}(e) \approx 0.497 \implies x * 0.434 \approx 0.497 \implies x \approx \dfrac{0.497}{0.434} \approx 1.145 .\)

It makes no difference what base you choose, but you must be consistent.
 

davehogan

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Many thanks for that explanation.
 

davehogan

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I'm having second thoughts about your "simple correct way" of solving this. How do you calculate "xloge(e) = 1.145?
 

JeffM

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I'm having second thoughts about your "simple correct way" of solving this. How do you calculate "xloge(e) = 1.145?
Oh my goodness

\(\displaystyle e^x = \pi \implies log_e(e^x) = log_e( \pi ),\ but\ log_e ( \pi) \approx 1.145.\)

\(\displaystyle So\ log_e(e^x) \approx 1.145 \implies x * log_e(e) \approx 1.145 \implies x * 1 \approx 1.145 \implies x \approx 1.145.\)

The point is that this is an equation. What does the right hand side of it equal, approximately in this case? What does the left hand side of it equal, exactly in this case?

The expression \(\displaystyle x * log_e(e)\) can equal any real number if it is not subject to any conditions.

But is subject to a condition, namely \(\displaystyle x * log_e(e) = log_e(\pi ).\)
 

davehogan

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OK.

Yours truly

A Dimwit.
 

JeffM

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