# Solving Logarithmic Equations

• Apr 22nd 2009, 06:17 PM
asimon2008
Solving Logarithmic Equations
Hello I'm a newbie. I need help with my homework. I have ADD bad. I don't understand this section Could someone please show how to work these problems?

In Ex 73-92, solve the logarithmic equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility.

81. 7 log4(0.6x)=12

83. ln sq. root(x+2)=1

89. ln(x+5)=ln(x-1)-ln(x+1)
• Apr 23rd 2009, 01:34 PM
Soroban
Hello, asimon2008!

Welcome aboard!

Quote:

Solve the logarithmic equation algebraically.
Round the result to three decimal places.

$(81)\;\;7\log_4(0.6x)\:=\:12$

We have: . $\log_4(0.6x) \:=\:\tfrac{12}{7}$

. . . . . . . $0.6x \;=\;4^{\frac{12}{7}}$

. . . . . . . $x \;=\; \frac{4^{|frac{12}{7}}}{0.6} \;=\;17.9453359 \;\approx\;17.945$

Quote:

$(83)\;\;\ln\sqrt{x+2} \:=\:1$

We have: . $\ln\sqrt{x+2} \:=\:1 \quad\Rightarrow\quad \sqrt{x+2} \;=\;e^1$

. . . . . . . $x+ 2 \;=\;e^2 \quad\Rightarrow\quad x \;=\;e^2-2\;\approx\;5.389$

Quote:

$(89)\;\;\ln(x+5)\:=\:\ln(x-1)-\ln(x+1)$

We have: . $\ln(x+5) \;=\;\ln\left(\frac{x-1}{x+1}\right) \quad\Rightarrow\quad x+5 \;=\;\frac{x-1}{x+1}$

. . . . . . . . $(x+5)(x+1) \:=\:x-1 \quad\Rightarrow\quad x^2 + 6x + 5 \;=\;x - 1$

. . . . . . . . $x^2 + 6x + 5 \:=\:0\quad\Rightarrow\quad (x+2)(x+3) \:=\:0$

Hence: . $x \;=\;\text{-}2,\:\text{-}3$

But both roots are extraneous . . . The equation has no solution.