More Complex Logarithm and advanced Math Questions

[jspen30]

Hello, was wondering if anyone could please help me with the following questions as for math I have been given a substitute teacher who is of little help.

Any help would be much appreciated, even if its just pointing me in the right direction

Equation 1:

Log (x-3) = 1 + Log 4 - Log x


Equation 2:

2-4a + 2 x 2-2a - 8 = 0


Equation 3:

Without using calculator, find the value of A that makes x = e^12 / 1-2e^12 a solution of the equation:

ln x - ln (ax+1) = 12




Thanks again Jake

 

[eumyang]

Equation 1:
[itex]\log (x-3) = 1 + \log 4 - \log x[/itex]
Rewrite 1 as a logarithm. Use the properties of logarithms to combine the logs on the right side into a single logarithm. Then "drop" the log from both sides, and solve.

Equation 2:
This is unreadable. What does the "-" right before the 4a and 2a mean? Are they subtractions? And is the "x" a variable or multiplication?

Equation 3:
[itex]\ln x - \ln (ax+1) = 12[/itex]
Similar to #1. Rewrite 12 as a natural logarithm. Use the properties of logarithms to combine the logs on the left side into a single logarithm. Then "drop" the ln from both sides, and solve.

 

[jspen30]

Thanks for your fast reply

Sorry Equation 2 is as follows, as the power symbol didnt work:

2^-4a + 2 x 2^-2a - 8 = 0

Thanks again

 

[Ray Vickson]

Thanks for your fast reply

Sorry Equation 2 is as follows, as the power symbol didnt work:

2^-4a + 2 x 2^-2a - 8 = 0

Thanks again

Again: is the "x" a variable, or is it a multiplication sign? If you mean multiplication, it would be much better to use an asterisk (*), like this: 2^(-4a) + 2*2^(-2a) - 8 = 0.

RGV

 

[PeterO]

Thanks for your fast reply

Sorry Equation 2 is as follows, as the power symbol didnt work:

2^-4a + 2 x 2^-2a - 8 = 0

Thanks again

Let U = 2^-2a; substitute, solve for u then find solutions for x