A couple logs: log(base7)(x+6)=0, etc

[klubbhead024]

For the life of me I can't figure these out.

1. log(base7)(x+6)=0
2. log(base6)(x^2-x)=1
3. logx+log(x-1)=1
4. 3logx-logx^2=2

My math book could not be more unclear about how to do these

 

[jonboy]

Hi klubbhead024!

\(\displaystyle \L \;log_7(x\,+\,6)\,=\,0\)

Raise both sides to a base of 7:\(\displaystyle \L \;x\,+\,6\,=\,1\)

You can finish that.

\(\displaystyle \L \;log_6(x^2-x)=1\)

Raise both sides to a base of 6 and solve the quadratic.

\(\displaystyle \L \;logx+log(x-1)=1\)

\(\displaystyle log[x(x\,-\,1)]\,=\,1\)

Raise the log to a base of 10:\(\displaystyle \;x^2\,-\,x\,-10\,=\0\)

Once again, solve the quadratic.

\(\displaystyle \L \;3logx-logx^2=2\)

\(\displaystyle \L \;log[x^3-\,x^2]=\,2\)

Raise both sides to a base of 10:\(\displaystyle \;x^3\,-\,x^2\,=\,2\)

\(\displaystyle x(x^2\,-\,x)\,-\,2\,=\,0\)

Solve for \(\displaystyle x\).

 

[Mrspi]

Re: A couple logs

For the life of me I can't figure these out.


4. 3logx-logx^2=2

My math book could not be more unclear about how to do these

Use the rules of logs to write the left-hand side as a single log.

Since log a<SUP>n</SUP> = n log a, 3 log x = log x<SUP>3</SUP>
log x<SUP>3</SUP> - log x<SUP>2</SUP> = 2

Since log (a/b) = log a - log b, log x<SUP>3</SUP> - log x<SUP>2</SUP> means log (x<SUP>3</SUP>/x<SUP>2</SUP>):

log (x<SUP>3</SUP>/x<SUP>2</SUP>) = 2

Or, log x = 2
Change to exponential form:
x = 10<SUP>2</SUP>