Solving equations and inequalities (logs and e)

**Danktoker** · March 30th 2010, 05:07 PM

I am not sure what to do with the following problems.
We are suppose to solve using algebraic method and give an exact answer in order to receive the most credit.

These are just some examples (there is alot more) but i feel that once i understand how to do these the rest will be some wat easier to figure out.

a)x^log(x)= x^3/100

b)3< 3e^(1/x^2+1)<3e

c)|10^x^2-10000|<9000

Once again thank you very much for any help. I really appreciate it.

-Gus

**Soroban** · March 30th 2010, 07:10 PM

Hello, Danktoker!

a) x^log() = x^3/100

Is that: . $x^{\log x} \:=\:\frac{x^3}{100}\;\;\text{ or }\;\;x^{\log x} \:=\:x^{\frac{3}{100}}$

b) 3 < 3e^(1/x^2+1) < 3e

Is that: . $3\:<\:3e^{\frac{1}{x^2} + 1} \:<\:3e\;\;\text{ or }\;\;3\:<\:3e^{\frac{1}{x^2+1}} \:<\:3e$

c) |10^x^2-10000| < 9000

We have: . $\left|10^{x^2} - 10,\!000\right| \;<\;9,\!000$

which equals: . $-9,\!000 \;<\;10^{x^2} - 10,\!000\;<\;9,\!000$

Add 10,000: . $1,\!000 \;<\; 10^{x^2} \;<\;19,\!000$

Take logs: . $\log(1000) \;<\;\log\left(10^{x^2}\right) \;<\;\log(19,\!000)$

. . Note: . $\log(19,\!000) \:=\:\log(19\cdot10^3) \:=\:\log(19) + \log(10^3)$

$\text{So we have: }\;\underbrace{\log\left(10^3\right)}_{\text{This is 3}} \;<\;x^2\underbrace{\log(10)}_{\text{This is 1}} \;<\;\log(19) + \underbrace{\log\left(10^3\right)}_{\text{This is 3}}$

. . . . . . . . . . . . . . $3 \quad<\quad x^2 \quad<\quad \log(19) + 3$

Therefore:. . . . . . . $\sqrt{3}\;\;<\;\;x\;\;<\;\;\sqrt{\log(19) + 3}$

**Danktoker** · March 30th 2010, 08:46 PM

for the first one it is the first one.

for the second one it is the second one.

Sory i dnot know how to copy what u did..

I hope that is clear.

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