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September 5th, 2014, 01:02 PM   #1
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solve exponential inequality

solve 3^(x+1/x)<=4
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September 5th, 2014, 02:13 PM   #2
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Step 1: Take the common logarithm or natural logarithm of each side.
Step 2: Use the properties of logarithms to rewrite the problem. Specifically, use Property which says
log(x^y) = y log x
Step 3: Divide each side by the logarithm.
Step 4: Use a calculator to find the decimal approximation of the logarithms.
Step 5: Finish solving the problem by isolating the variable...
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September 5th, 2014, 02:36 PM   #3
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I don't agree with step four.
In step 5 you must take care to note whether $x$ is greater or less than zero.
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September 5th, 2014, 02:54 PM   #4
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$\displaystyle 3^{x+\frac1x}\le4$

so we must have

$\displaystyle \frac{x^2+1}{x}\le\log_34$

Since $\displaystyle \frac{x^2+1}{x}$ may be written as $\displaystyle \frac{(x-1)^2}{x}+2$

the minimum of $\displaystyle \frac{x^2+1}{x}$ is 2, for $\displaystyle x>0$.

Hence the original statement is only true for x < 0.
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September 5th, 2014, 03:13 PM   #5
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Originally Posted by mared View Post
solve 3^(x+1/x)<=4
Please use thread titles that indicate something about the question you are asking others to do for you. A thread title such as "solve" is useless to those viewing the thread listings.
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