# Natural and common logarithms inequalities?

• May 29th 2011, 11:57 AM
homeylova223
Natural and common logarithms inequalities?
I have two problems one is a common logarithm and the other is a natural logarithm.

1. Use natural logarithms to solve each inequality
x^(2/3) greater than to 27.6

(2/3) ln x greater than ln 27.6
And I am stuck right here because if I put ln(x) in my calculator
I get 2.3025 but If proceed with this I get the wrong answer.

2. This next problem I must solve using common logarithms

log x^(6) greater than 1 I am not sure how to start this one(Giggle)
• May 29th 2011, 12:13 PM
Quacky
Quote:

Originally Posted by homeylova223
I have two problems one is a common logarithm and the other is a natural logarithm.

1. Use natural logarithms to solve each inequality
x^(2/3) greater than to 27.6

(2/3) ln x greater than ln 27.6
And I am stuck right here because if I put ln(x) in my calculator
I get 2.3025 but If proceed with this I get the wrong answer.

2. This next problem I must solve using common logarithms

log x^(6) greater than 1 I am not sure how to start this one(Giggle)

$\displaystyle x^{\frac{2}{3}}>27.6$
Take logs of both sides:

$\displaystyle \displaystyle Log(x^{\frac{2}{3}})>Log(27.6)$

$\displaystyle \frac{2}{3}Log(x)>Log(27.6)$

$\displaystyle Log(x)>\frac{3}{2}Log(27.6)$

$\displaystyle Log(x)>Log(27.6^{\frac{3}{2}})$

$\displaystyle x>27.6^{\frac{3}{2}}$

Try b) yourself - the method is similar!
• May 30th 2011, 09:19 AM
homeylova223
For my second problem would I do this

log x^(6) greater than 1
6 log x greater than log 1 (Then how would I proceed?)
• May 30th 2011, 09:28 AM
e^(i*pi)
Quote:

Originally Posted by homeylova223
For my second problem would I do this

log x^(6) greater than 1
6 log x greater than log 1 (Then how would I proceed?)

The original version of the question (in the OP) states that $\displaystyle \log(x^6) > 1$. If this is the case you can use the basic definition of a logarithm: $\displaystyle x = 10^{\log_{10}(x)}$
• May 30th 2011, 02:32 PM
homeylova223
Hmm I am still kind of confused on how to solve it?
• May 30th 2011, 02:39 PM
e^(i*pi)
$\displaystyle 10^{\log(x^6)} > 10^{1} \rightarrow x^6 > 10$
• May 30th 2011, 04:30 PM
Soroban
Hello, homeylova223!

Quote:

$\displaystyle \text{1. Use natural logarithms to solve the inequality: }\:x^{\frac{2}{3}} \:> \:27.6$

Take logs: .$\displaystyle \ln\left(x^{\frac{2}{3}}\right) \:>\:\ln(27.6)$

. . . . . . . . . .$\displaystyle \tfrac{2}{3}\ln(x) \:>\:\ln(27.6)$

. . . . . . . . . . . $\displaystyle \ln(x) \:>\:\tfrac{3}{2}\ln(27.6) \:=\:4.976723659$

. . . . . . . . . . . . . .$\displaystyle x \:>\:e^{4.976723659}$

. . . . . . . . . . . . . .$\displaystyle x \:>\:144.9985379$

Quote:

$\displaystyle \text{2. Solve using common logarithms: }\:\log(x^6)\: > \: 1$

We have: .$\displaystyle \log(x^6) \:>\:\log(10)$

. . . Then: . . . . $\displaystyle x^6 \:>\:10$

Therefore: . . . .$\displaystyle x \:>\:\sqrt[6]{10}$

. . . . . . . . . . . . $\displaystyle x \:>\:1.467799268$

• Jun 9th 2011, 11:12 AM
HallsofIvy
It is important to note that the logarithm, with any positive base, is an increasing function: if a> b then log(a)> log(b).