I hope I am correctly following the rules on new posts, because this is a new post. What is log5 square root of 125 minus log2 square root of 2? I took the square root of 125, or approximately 11.18, and converted it to log10, or 1.048. I took the square root of 2 or approximately, 1.414, and divided 1.048 by 1.414. I get 1.35. I know the correct answer is one, but I am not sure if I made a rounding or approximation error? Thanks for your help very much, 0313
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College Algebra Logarithmic Equation Help
- Thread starter 0313phd
- Start date
Is this what you mean?:
\(\displaystyle log_{5}\sqrt{125}-log_{2}\sqrt{2}\)
If so, change of base:
\(\displaystyle \frac{log(\sqrt{5^{3}})}{log(5)}-\frac{log(\sqrt{2})}{log(2)}\)
\(\displaystyle \frac{\frac{3}{2}log(5)}{log(5)}-\frac{\frac{1}{2}log(2)}{log(2)}\)
Now, finish?.
\(\displaystyle log_{5}\sqrt{125}-log_{2}\sqrt{2}\)
If so, change of base:
\(\displaystyle \frac{log(\sqrt{5^{3}})}{log(5)}-\frac{log(\sqrt{2})}{log(2)}\)
\(\displaystyle \frac{\frac{3}{2}log(5)}{log(5)}-\frac{\frac{1}{2}log(2)}{log(2)}\)
Now, finish?.
I am lost after this first line of your response. I did change the log5 square root of 125 to base 10 for the square root of 125, and got 11.18 rounded. Then I changed the base of 2, square root to base 10. I got .1505, rounded. Then I divided 11.18 by .1505, rounded. I think what you did is to put the exponents up from, and cancell out the logs, so you could subtract the exponents to get 1. Am I seeing what you did correctly?
0313phd said:I hope I am correctly following the rules on new posts, because this is a new post. What is log5 square root of 125 minus log2 square root of 2? I took the square root of 125, or approximately 11.18, and converted it to log10, or 1.048. I took the square root of 2 or approximately, 1.414, and divided 1.048 by 1.414. I get 1.35. I know the correct answer is one, but I am not sure if I made a rounding or approximation error? Thanks for your help very much, 0313
Hmmm....I think this problem can be done without a calculator....
125 is 5[sup:egxkzkk9]3[/sup:egxkzkk9]
sqrt(125), then is (5[sup:egxkzkk9]3[/sup:egxkzkk9])[sup:egxkzkk9](1/2)[/sup:egxkzkk9], or 5[sup:egxkzkk9](3/2)[/sup:egxkzkk9] using the rule of exponents which says that when you raise a power to a power, you multiply the exponents.
log[sub:egxkzkk9]5[/sub:egxkzkk9] 5[sup:egxkzkk9](3/2)[/sup:egxkzkk9] is, by definition, the power to which 5 must be raised to produce 5[sup:egxkzkk9](3/2)[/sup:egxkzkk9]. So, log[sub:egxkzkk9]5[/sub:egxkzkk9] 5[sup:egxkzkk9](3/2)[/sup:egxkzkk9] = 3/2
Apply the same kind of reasoning to log[sub:egxkzkk9]2[/sub:egxkzkk9] sqrt(2). Remember that sqrt(2) can be written as 2[sup:egxkzkk9](1/2)[/sup:egxkzkk9]. So, log[sub:egxkzkk9]2[/sub:egxkzkk9] 2[sup:egxkzkk9](1/2)[/sup:egxkzkk9] = 1/2
log[sub:egxkzkk9]5[/sub:egxkzkk9] sqrt(125) - log[sub:egxkzkk9]2[/sub:egxkzkk9] sqrt(2) = (3/2) - (1/2)
You shouldn't need a calculator to produce an EXACT answer.
J
JeffM
Guest
0313phd said:I am lost after this first line of your response. I did change the log5 square root of 125 to base 10 for the square root of 125, and got 11.18 rounded. Then I changed the base of 2, square root to base 10. I got .1505, rounded. Then I divided 11.18 by .1505, rounded. I think what you did is to put the exponents up from, and cancell out the logs, so you could subtract the exponents to get 1. Am I seeing what you did correctly?
First, you followed the rules for posting quite nicely.
Galactus and mrspi have shown you two distinct processes for solving your specific problem, but I fear they have not got to the root of your trouble.
What you originally did was to approximate 125[sup:2xkxzh8e](1/2)[/sup:2xkxzh8e], getting 11.18. Good approximation. Then you approximated log[sub:2xkxzh8e]10[/sub:2xkxzh8e](11.18) to get 1.048. Another good
approximation. Calculators will give you great approximations of whatever you ask them to approximate. You got a good approximation for
log[sub:2xkxzh8e]10[/sub:2xkxzh8e][125[sup:2xkxzh8e](1/2)[/sup:2xkxzh8e]]. BUT that is not what the problem asked about. It asked about log[sub:2xkxzh8e]5[/sub:2xkxzh8e][125[sup:2xkxzh8e](1/2)[/sup:2xkxzh8e]]. You approximated
2[sup:2xkxzh8e](1/2)[/sup:2xkxzh8e] at 1.414.Another triumph of the calculator. Then you (somehow) calculated that log at 1.35. By inspection, that cannot be the right answer. 1.414 is
less than both 2 and 10 so the log of 1.414 to either base must be less than 1.
I suspect you have two sources of confusion. First, the calculator tempts you to start taking approximations before you have done any analysis to see WHETHER you can get an EXACT answer or WHAT you should be approximating. Both mrspi's answer and galactus's answer first analyze by using the properties of logs and thereby determine that an exact answer is possible. Second, you do not seem to have grasped fully the concept of a logarithm and the significance of the base within that concept.
Hello, 0313phd!
From the nature of the problem, we are expected to do this without a calculator.
\(\displaystyle \log_5(\sqrt{125}) \;=\;\log_5(125)^{\frac{1}{2}} \;=\;\log_5(5^3)^{\frac{1}{2}} \;=\;\log_5(5^{\frac{3}{2}}) \;=\;\tfrac{3}{2}\underbrace{\log_5(5)}_{\text{This is 1}} \;=\;\tfrac{3}{2}\)
\(\displaystyle \log_2(\sqrt{2}) \;=\;\log_2(2^{\frac{1}{2}}) \;=\;\tfrac{1}{2}\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\tfrac{1}{2}\)
\(\displaystyle \text{Therefore: }\;\log_5(\sqrt{125}) - \log_2(\sqrt{2}) \;=\;\tfrac{3}{2} - \tfrac{1}{2} \;=\;1\)
\(\displaystyle \text{Evaluate: }\:\log_5(\sqrt{125}) - \log_2(\sqrt{2})\)
From the nature of the problem, we are expected to do this without a calculator.
\(\displaystyle \log_5(\sqrt{125}) \;=\;\log_5(125)^{\frac{1}{2}} \;=\;\log_5(5^3)^{\frac{1}{2}} \;=\;\log_5(5^{\frac{3}{2}}) \;=\;\tfrac{3}{2}\underbrace{\log_5(5)}_{\text{This is 1}} \;=\;\tfrac{3}{2}\)
\(\displaystyle \log_2(\sqrt{2}) \;=\;\log_2(2^{\frac{1}{2}}) \;=\;\tfrac{1}{2}\underbrace{\log_2(2)}_{\text{This is 1}} \;=\;\tfrac{1}{2}\)
\(\displaystyle \text{Therefore: }\;\log_5(\sqrt{125}) - \log_2(\sqrt{2}) \;=\;\tfrac{3}{2} - \tfrac{1}{2} \;=\;1\)