Logarithms

**currypuff** · March 4th 2008, 07:44 AM

Please help me solve the following problems:

1.) $log_4(3x-2)^{1/5} = 1/10$

2.) $ln(x) + ln(x+6) = 0.5(ln 9)$

3.) $30,000 = \frac {50,500}{1 + 100e^{-0.7x}}$

Your help is greatly appreciated.

**colby2152** · March 4th 2008, 08:08 AM

Originally Posted by currypuff

Please help me solve the following problems:

1.) $log_4(3x-2)^{1/5} = 1/10$

2.) $ln(x) + ln(x+6) = 0.5(ln 9)$

3.) $30,000 = \frac {50,500}{1 + 100e^{-0.7x}}$

Your help is greatly appreciated.

1.) $log_4(3x-2)^{1/5} = 1/10$

$(3x-2)^{1/5}=\left(4\right)^{\frac{1}{10}}$

$(3x-2)=\left(4\right)^{\frac{5}{10}}$

$3x-2=\sqrt{4}$

$(3x-2)^2=4$

$9x^2-12x+4=4$

$9x^2=12x$

$x \ne 0$

$9x=12$

$x=\frac{12}{9}$

**Soroban** · March 4th 2008, 11:18 AM

Hello, currypuff!

$1)\;\;\log_4(3x-2)^{\frac{1}{5}} \;=\; \frac{1}{10}$

We have: . $\frac{1}{5}\log_4(3x-2) \:=\:\frac{1}{10}$

Multiply by 5: . $\log_4(3x-2)\:=\:\frac{1}{2}$

Exponentiate: . $3x-2\;= \;4^{\frac{1}{2}}\quad \Rightarrow\quad 3x-2 \;=\;2$

Therefore:. $3x\;=\;4\quad\Rightarrow\quad\boxed{x \:=\:\frac{4}{3}}$

$2)\;\;\ln(x) + \ln(x+6) \:= \:\frac{1}{2}\ln 9$

We have: . $\ln[x(x+6)] \:=\:\ln(9^{\frac{1}{2}})\quad\Rrightarrow\quad \ln(x^2+6x) \:=\:\ln(3)<br />$

"Un-log" both sides: . $x^2-6x\:=\:3\quad\Rightarrow\quad x^2 -6x - 3\:=\:0$

Quadratic Formula: . $x \;=\;\frac{-6 \pm\sqrt{48}}{2} \;=\;-3 \pm2\sqrt{3}$

The negative root is extraneous.

Therefore: . $\boxed{x \;=\;2\sqrt{3} - 3}$

$3)\;\;\frac{50,500}{1 + 100e^{-0.7x}} \;=\;30,000$

Divide by 50,500: . $\frac{1}{1 + 100e^{-0.7x}} \:=\:\frac{60}{101}$

"Cross-multiply": . $101 \:=\:60 + 6000e^{-0.7x} \quad\Rightarrow\quad 41 \;=\;6000^{-0.7x}$

Multiply by $e^{0.7x}\!:\;\;41e^{0.7x}\;=\;6000 \quad\Rightarrow\quad e^{0.7x} \;=\;\frac{6000}{41}$

Therefore: . $0.7x \;=\;\ln\left(\frac{6000}{41}\right) \quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{0.7}\ln\left(\frac{6000}{41}\right)}<br /> <br />$

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