Using an Exponent to Undo Log

[Jason76]

Does this reasoning look ok?

Note:\(\displaystyle \log(h) = \log_{10}(h)\) where \(\displaystyle h = \) a number (not sure which kind)

\(\displaystyle \log x - \dfrac{1}{3} \log 8 = \log 7\)

\(\displaystyle 10^{\log x} - 10^{\frac{1}{3}\log 8} = 10^{\log 7}\)

\(\displaystyle x - 10^{\frac{1}{3}}(8) = 7\) - What is \(\displaystyle 10\) to the \(\displaystyle 1/3\) power?

The book was using some method using the power, product, etc.. rules of log to solve the equation. In the end, the correct answer should come out to \(\displaystyle x = 14\)

 

[HallsofIvy]

Does this reasoning look ok?

Note:\(\displaystyle \log(h) = \log_{10}(h)\) where \(\displaystyle h = \) a number (not sure which kind)

\(\displaystyle \log x - \dfrac{1}{3} \log 8 = \log 7\)

\(\displaystyle 10^{\log x} - 10^{\frac{1}{3}\log 8} = 10^{\log 7}\)

\(\displaystyle x - 10^{\frac{1}{3}}(8) = 7\) - What is \(\displaystyle 10\) to the \(\displaystyle 1/3\) power?

The book was using some method using the power, product, etc.. rules of log to solve the equation. In the end, the correct answer should come out to \(\displaystyle x = 14\)

You are making several errors here. Taking 10 to the power of each side is legal but on the left you get \(\displaystyle 10^{\log(x)- \frac{1}{3}\log(8)}\) which is NOT the same as \(\displaystyle 10^{\log(x)}- 10^{\frac{1}{3}\log(8)}\). That is, \(\displaystyle 10^{a- b}\) is NOT the same as \(\displaystyle 10^a- 10^b\).

 

[Jason76]

You are making several errors here. Taking 10 to the power of each side is legal but on the left you get \(\displaystyle 10^{\log(x)- \frac{1}{3}\log(8)}\) which is NOT the same as \(\displaystyle 10^{\log(x)}- 10^{\frac{1}{3}\log(8)}\). That is, \(\displaystyle 10^{a- b}\) is NOT the same as \(\displaystyle 10^a- 10^b\).

So what would be the next step after this (using the correct method)?

 

[JeffM]

Does this reasoning look ok?

Note:\(\displaystyle \log(h) = \log_{10}(h)\) where \(\displaystyle h = \) a number (not sure which kind) I think all that is being said here is that

\(\displaystyle log(z)\ MEANS\ log_{10}(z).\)

By the way, when I was a kid, that was standard. ln meant log to the base e. log meant log to the base 10. If some other base was intended, it was explicitly shown. I gather that such a convention is no longer standard. Back then, you used logs to do computations that modern calculators do in nanoseconds so logs were used frequently.

\(\displaystyle \log x - \dfrac{1}{3} \log 8 = \log 7\)

\(\displaystyle 10^{\log x} - 10^{\frac{1}{3}\log 8} = 10^{\log 7}\) I have explained before that this kind of transformation is seldom very helpful.

\(\displaystyle x - 10^{\frac{1}{3}}(8) = 7\) - What is \(\displaystyle 10\) to the \(\displaystyle 1/3\) power?

The book was using some method using the power, product, etc.. rules of log to solve the equation. In the end, the correct answer should come out to \(\displaystyle x = 14\)

For goodness sake, stop using this seldom useful transformation and use the laws of logs that were INVENTED to make this sort of thing easy.

\(\displaystyle log(x) - \dfrac{1}{3}log(8) = log(7) \implies a - b = c \implies a - c = b \implies\)

\(\displaystyle log(x) - log(7) = \dfrac{1}{3}log(8) \implies\)

\(\displaystyle log\left(\dfrac{x}{7}\right) = log\left(8^{(1/3)}\right).\)

So what next?

Things like logs and the laws of logs are TOOLS. You need to use them. Eliminating them as quickly as you can means that you are not using the tools that God and Napier gave you.

 

[DrPhil]

\(\displaystyle \log x - \dfrac{1}{3} \log 8 = \log 7\)

\(\displaystyle \dfrac{1}{3}\log (8) = \log(8^{1/3}) = \log(2)\)

\(\displaystyle \log x - \log 2 = \log \dfrac{x}{2} = \log 7 \)

\(\displaystyle \dfrac{x}{2} = 7\)

\(\displaystyle x = 14 \)

 

[Jason76]

\(\displaystyle \dfrac{1}{3}\log (8) = \log(8^{1/3}) = \log(2)\)

\(\displaystyle \log x - \log 2 = \log \dfrac{x}{2} = \log 7 \)

\(\displaystyle \dfrac{x}{2} = 7\)

\(\displaystyle x = 14 \)

Makes sense, but I still never got an answer from anyone on how to do math when making ALL logs a power of 10 (or some other base). Another situation might come up, that is not related to log equations where that might come up.

\(\displaystyle 10^{\log(x)- \frac{1}{3}\log(8)} = 10^{\log7}\)

 

[JeffM]

Makes sense, but I still never got an answer from anyone on how to do math when making ALL logs a power of 10 (or some other base). Another situation might come up, that is not related to log equations where that might come up.

When it comes up (if it ever does), you must use this transformation that you love and the laws of exponents. In the meantime, you are NOT learning how to use logs.

Please do not get angry, but you seem insistent on using a very small number of tools. You cannot build a skyscraper with a hammer and a saw, no matter how useful those tools may be on the right occasion. A lot of math is learning to use a varied and ultimately very large set of tools cleverly. When you refuse to learn when specific tools are useful, you lose 80% of what math is about. I do not say this as a mathematician. My academic field was European history, and my career was partly analytical and partly managerial. I can assure you that in the analytic part you want as many tools as possible. (As for the managerial part, math is little help though an intelligent study of history may be: people do not follow rules the way numbers do.)

 

[Jason76]

When it comes up (if it ever does), you must use this transformation that you love and the laws of exponents. In the meantime, you are NOT learning how to use logs.

Please do not get angry, but you seem insistent on using a very small number of tools. You cannot build a skyscraper with a hammer and a saw, no matter how useful those tools may be on the right occasion. A lot of math is learning to use a varied and ultimately very large set of tools cleverly. When you refuse to learn when specific tools are useful, you lose 80% of what math is about. I do not say this as a mathematician. My academic field was European history, and my career was partly analytical and partly managerial. I can assure you that in the analytic part you want as many tools as possible. (As for the managerial part, math is little help though an intelligent study of history may be: people do not follow rules the way numbers do.)

Ok that makes sense.

Going back to the "law of exponents" thing: If you add exponents, then you multiply bases, but if you subtract exponents, then you would divide the bases. Is that right?

 

[lookagain]

Note:\(\displaystyle \log(h) = \log_{10}(h)\) where \(\displaystyle h = \) a number (not sure which kind)

h belongs to the set of positive real numbers.\(\displaystyle \ \ \ \)

Going back to the "law of exponents" thing: If you add exponents, then you multiply bases, but if you subtract exponents, then you would divide the bases. Is that right?

If two numbers with the same bases are multiplied together, then the result is equal to that base raised to the sum of those exponents. \(\displaystyle \ \ \ \ \ \ \ \ \ \) If two numbers with the same bases are divided, then the result is equal to that base raised to the difference** of those exponents. \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \) **(exponent of the dividend minus the exponent of the divisor)

 

[Jason76]

\(\displaystyle \dfrac{1}{3}\log (8) = \log(8^{1/3}) = \log(2)\)

\(\displaystyle \log x - \log 2 = \log \dfrac{x}{2} = \log 7 \)

\(\displaystyle \dfrac{x}{2} = 7\)

\(\displaystyle x = 14 \)

\(\displaystyle \log x - \log 2 = \log 7\)

\(\displaystyle \log \dfrac{x}{2} = \log 7\)

\(\displaystyle \log \dfrac{x}{2} = \log 7\) - This step needs to be exponanted (Is that the right word?)

\(\displaystyle 10^{\log \frac{x}{2}} = 10^{\log 7}\)

\(\displaystyle \dfrac{x}{2} = 7\) - But of course, that leads to the same answer.

\(\displaystyle x = 14 \)

 

[Jason76]

An example of exponanted stuff:

\(\displaystyle e^{\ln (2 + y^{2)}} = e^{\ln(4 + x^{2}) + C}\)

\(\displaystyle 2 + y^{2} = (4 + x^{2}) e^{C}\) Since adding, then multiplying comes in.

But with the other example on this page, some subtracting was in there.

 

[JeffM]

\(\displaystyle \log x - \log 2 = \log 7\)

\(\displaystyle \log \dfrac{x}{2} = \log 7\)

\(\displaystyle \log \dfrac{x}{2} = \log 7\) - This step needs to be exponanted (Is that the right word?)

\(\displaystyle 10^{\log \frac{x}{2}} = 10^{\log 7}\)

\(\displaystyle \dfrac{x}{2} = 7\) - But of course, that leads to the same answer.

\(\displaystyle x = 14 \)

Lookagain's answer to your question about exponents should have been sufficient.

You are, I think, adding an extra and unnecessary step

\(\displaystyle Given\ a, b, c > 0\ and\ a \ne 1,\ log_a(b) = log_a(c) \iff b = c.\) This is a general and very useful rule.

You can go straight from \(\displaystyle log\left(\dfrac{x}{2}\right) = log(7)\)

to \(\displaystyle \dfrac{x}{2} = 7\) without the intervening exponentiation step.

 

[Jason76]

Lookagain's answer to your question about exponents should have been sufficient.

You are, I think, adding an extra and unnecessary step

\(\displaystyle Given\ a, b, c > 0\ and\ a \ne 1,\ log_a(b) = log_a(c) \iff b = c.\) This is a general and very useful rule.

You can go straight from \(\displaystyle log\left(\dfrac{x}{2}\right) = log(7)\)

to \(\displaystyle \dfrac{x}{2} = 7\) without the intervening exponentiation step.

In regard to Lookagain's answer, he didn't write down the exponation step. That could confuse some people.

 

[JeffM]

In regard to Lookagain's answer, he didn't write down the exponation step. That could confuse some people.

I will not put words into lookagain's mouth; he is perfectly able to speak for himself.

All I shall say is that:

(1) I read his comment as referring to your question about exponents, not your more general question, and saw no need to elaborate what I believe to have been a sufficient answer; and

(2) the exponentiation step is, in my opinion, unnecessary.

Now if you found his explanation about exponents unclear or incomplete, please let us know where one of us can add something.