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log5(3x+9)-2
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Express as a logarithm of a single number or expression:
1. 5log4^p+log4^q 2. log10^x-4 log10^y 3. 4log3^A-1/2 log3^B 4.
Top answer:
Sure! Let's solve each expression step by step: 1. To express 5log4^p + log4^q as a logarithm of a
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Given that log5 3=0.683 and log5 7=1.209.without using calculator,evaluate
(a) log5 1.4 b)log7 75 c) log3 125
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Did you notice that 1.4 = 7/5 ? log<sub>5</sub> 1.4 = log<sub>5</sub> (7/5) = log<sub>5</sub> 7 -
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solve the logarithmic equation . express solution in exact form
log5(x-9)+log5(x+4)=1+log5(x-5)
Top answer:
1 = log5(5), so we have log5(x-9)+log5(x+4)=log5(5)+log5(x-5) log5[(x-9)(x+4)] = log5[5(x-5)] raise
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log5(x)-log5(x+4)=-1 (the 5s are in the base of the log)
I need to find x I know you can combine them into log5(x/x+4)= -1 but i
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log5(x)-log5(x+4)=-1 log5[ x/(x+4) ] = -1 now use your definition of logs x/(x+4) = 5^-1 = 1/5 5x =
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Log6(6^9)=???
Answer: 9 Ine^-3x=??? Answer: -3x Use log5(2) = 0.4307 and log5(3) = 0.6826 to approximate the value of log5(12)
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Log6(6^9)=??? 9 log6(6) = 9*1 Answer: 9 Right Ine^-3x=??? Answer: -3x Right Use log5(2) = 0.4307 and
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I cannot figure out how to work this one. I have tried working it like the example in the book but cannot come up with one of
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To solve this problem, we can use the logarithmic property that says: log(base a) (b^c) = c *
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How do you graph the following logs?
f(x)=log5 (x-2) f(x)=log5 x-2 f(x)= log5 x f(x)=log5 (x+2)
Top answer:
use f(x)= log<sub>5</sub> x as your reference graph. You should know what your standard log function
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using logarithms to solve exponential equations.
5^1+x = 2^1-x I need exact numbers. I did one on my own already. 5^x-1 = 9 5^x-1
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come on - simple algebra: (1+x)/(1-x) = .4306765581 1+x = .4306765581(1-x) Think you can handle it
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Use log5(2)=0.4307 and log5(3)=0.6826 to approximate the value of log5=54.
According to the example in my book I would divide
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Solve pls
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using logarithms to solve exponential equations.
5^1+x = 2^1-x I need exact numbers. I did one on my own already. 5^x-1 = 9 5^x-1
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I really need help
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