log(5)^2 + log(2)log(50)

[chocopax750]

Could someone help me solve this question?
Evaluate log(5)^2 + log(2)log(50).
Without calculator of course.

 

[chocopax750]

the key properties of logs

\(\displaystyle \log(a)+\log(b)=\log(ab)\)

\(\displaystyle \log(a)-\log(b)=\log(\frac{a}{b})\)

\(\displaystyle a\log(b)=\log(b^a)\)

see if you can use these to make any headway. Yell back if you still need help.

I don't really see the link. I know about the properties of logs, but somehow, I just couldn't apply them to the multiplication of logs.

 

[chocopax750]

using logs base 10

\(\displaystyle \log(2)\log(50)=\log(2^{\log(50)})\)

\(\displaystyle \log(5)^2=\log(5^{\log(5)})\)

\(\displaystyle \log(5)^2 + \log(2)\log(50)=\log(2^{\log(50)})+\log(5^{\log(5)})\)

\(\displaystyle \log(5)^2 + \log(2)\log(50)=\log\left(2^{\log(50)}\cdot5^{\log(5)}\right)=\log\left(2^{\log(5)+\log(10)}\cdot 5^{\log(5)}\right)\)

\(\displaystyle \log(5)^2 + \log(2)\log(50)=\log\left((2\cdot 5)^{\log(5)}\cdot 2^{\log(10)}\right)=\log\left(10^{\log(5)}\cdot 2^1\right)=\log\left(5\cdot 2)\right)=\log(10)=1\)

check this over for accuracy. I'm known to make mistakes in this sort of lengthy algebra.

The answer is correct, but I don't get the first step. How did you manage to turn the question into an equation?

 

[chocopax750]

I understand everything except for the connection between the fourth and fifth step.

 

[chocopax750]

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[chocopax750]

Thanks man! You cleared my mind over this question

 

[lookagain]

Could someone help me solve this question?

Evaluate log(5)^2 + log(2)log(50).

Without calculator of course.

chocopax750, this problem is geared toward the following method of solution:


\(\displaystyle [log(5)]^2 \ + \ [log(2)][log(50)] \ = \)

\(\displaystyle [log(5)]^2 \ + \ [log(10/5)][log(10*5)] \ = \)

\(\displaystyle [log(5)]^2 \ + \ [log(10) - log(5)][log(10) + log(5)] \ = \)

\(\displaystyle [log(5)]^2 \ + \ [1 - log(5)][1 + log(5)] \ = \)

\(\displaystyle [log(5)]^2 \ + \ 1 \ - \ [log(5)]^2 \ =\)

\(\displaystyle \boxed{ \ 1 \ }\)

 

[Dale10101]

Two cents

Thanks man! You cleared my mind over this question

Wow! … one unusual problem (how often do you see the product of two logs in an expression … not too blink’n often) and two entirely different solution methods!

I learned about four different transformations that are obvious in hindsight but seemed insurmountable looking forward … exciting (and, alas, depressing).

But hey, Nova recently featured an interview with a top administrator of the NSA who said every year just before summer break they send to high school students a list of problems that their analysts are having trouble with and, he says, they never fail to be surprised by the positive results brought forth by fresh perspectives.

This site is a valuable resource and “kuel” … (yes, of course, how nerdy can I get, fortunately I use an alias):cool:

 

[pka]

Could someone help me solve this question?
Evaluate log(5)^2 + log(2)log(50).
Without calculator of course.

Although I do not like to do it, I accept here \(\displaystyle \log(x)=\log_{10}(x) \)