Properties of Logarithms and solving log equations

[bluegirl]

I have these problems to do for homework for my algebra 2 class on the properties of logarithms. However, I have no idea how to solve them.

1. log 7 + log (n-2)=log6n

2. log 2(x) + 5=8-log 2 (x+7)

You would add log 2(x+7) to both sides and subtract 5 to both sides and get:

. . .log 2(x) + log 2(x+7)=3

Since the equation is an addition of logs with the same base, you can multiply the x and x+7, and then change it to exponential form and get:

. . .x(x+7)=2^3
. . .x^2+7x=8
. . .x^2+7x-8=0
. . .(x-1) (x+8)=0
. . .x-1=0 or x+8=0
. . .x=1 or x= -8

Is this answer right?

 

[stapel]

2. log 2(x) + 5=8-log 2 (x+7)

. . .x=1 or x = -8

Is this answer right?

You can check the solution to any "solving" problem by plugging it back in:

. . .x = 1:
. . . . .log₂([1]) + log₂([1] - 7)
. . . . .= log₂(1) + log₂(-6)
. . . . .=....

But you can't take the log of a negative, so this can't work.

What happens when you check the other solution?

Note: You changed the sign in the middle of the exercise, going from "x - 7" to "x + 7". Since I cannot see your text, I cannot tell which is correct, if either. But an error here would naturally influence the answer.

1. log 7 + log (n-2)=log6n

I will guess that the above means the following:

. . . . .log₁₀(7) + log₁₀(n - 2) = log₁₀(6n)

This exercise works just like the other one.

Eliz.

 

[soroban]

Hello, bluegirl!

\(\displaystyle 1.\;\log(7)\,+\,\log(n\,-\,2)\:=\:\log(6n)\)

Combine logs: \(\displaystyle \:\log[7(n\,-\,2)] \:=\:\log(6n)\)

"Un-log" both sides: \(\displaystyle \:7(n\,-\,2)\:=\:6n\)

Solve for \(\displaystyle n:\;\;7n\,-\,14\:=\:6n\;\;\Rightarrow\;\;n \,=\,14\)

\(\displaystyle 2.\;\log_2(x)\,+\,5\:=\:8\,-\,\log_2(x\,+\,7)\)

You would add \(\displaystyle \log_2(x+7)\) to both sides and subtract 5 to both sides and get:

. . .\(\displaystyle \log_2(x)\,+\,\log_2(x\,-\,7)\:=\:3\;\) Right!

Since the equation is an addition of logs with the same base,
you can multiply the \(\displaystyle x\) and \(\displaystyle x\,+\,7\),
and then change it to exponential form and get:

. . .\(\displaystyle x(x\,+\,7)\:=\:2^3\;\) Yes!

. . .\(\displaystyle x^2\,+\,7x\:=\:8\)

. . .\(\displaystyle x^2\,+\,7x\,-\,8\:=\:0\)

. . .\(\displaystyle (x\,-\,1) (x\,+\,8)\:=\:0\)

. . .\(\displaystyle x\,-\,1\:=\:0\,\) or \(\displaystyle \,x\,+\,8\:=\:0\)

. . .\(\displaystyle x\,=\,1\,\) or \(\displaystyle \,x\,=\,-8\;\;\) Good!

Is this answer right? . Not quite

Always check for extraneous roots.

In this problem, \(\displaystyle x\,=\,-8\) is extraneous.

The only solution is: \(\displaystyle \,x\,=\,1\)

 

[bluegirl]

oh, okay, so whenever you get an extraneous roots like I did in the last problem, you don't use it. You only use the solution that is not negative.

 

[stapel]

You only use the solution that is not negative.

Sometimes negative solutions are fine. You have to check the solution values in the original equation to check. For instance, "x = -8" for "log₂(-x) = 3" would be correct.

Don't just arbitrarily discard any and all negative solutions. Check first.

Eliz.

 

[bluegirl]

okay, so I should alawys check my solution before I discard, whether its positive or negative.