Solving for X: The Logarithmic Equations

[Alykayy]

Homework Statement

Solve for X

logx-log(x+11) = -1

and

log₄x-log₄(x+15) = -1

Homework Equations

The Attempt at a Solution

log x - log (x+11) = -1
log (x/x+11) = -1

I don't know how to solve for X after this point

log₄x-log₄(x+15) = -1
log₄ x/(x+15) = -1I don't know how to get the X out of the log to solve for X

 

[Mark44]

Homework Statement

Solve for X

logx-log(x+11) = -1

and

log₄x-log₄(x+15) = -1

Homework Equations

The Attempt at a Solution

log x - log (x+11) = -1
log (x/x+11) = -1

I don't know how to solve for X after this point

log₄x-log₄(x+15) = -1
log₄ x/(x+15) = -1I don't know how to get the X out of the log to solve for X

The relationship you need for both problems is this one:
log_a(x) = y is equivalent to x = a^y.

For your first problem, log means log₁₀ (lob base 10).

 

[Ray Vickson]

Homework Statement

Solve for X

logx-log(x+11) = -1

and

log₄x-log₄(x+15) = -1

Homework Equations

The Attempt at a Solution

log x - log (x+11) = -1
log (x/x+11) = -1

I don't know how to solve for X after this point

log₄x-log₄(x+15) = -1
log₄ x/(x+15) = -1I don't know how to get the X out of the log to solve for X

What "base" of logs is used in the first question?

Anyway, you certainly cannot have what you wrote, which was
$$\log \left( \frac{x}{x} + 11 \right) = -1$$
which gives $\log(12) = -1$. Did you mean
$$\log\left( \frac{x}{x+11} \right) = -1?$$
If so, use parentheses, like this: log(x/(x+11)) = -1. At his point it matters what base you are using for log.

For a given base $b$, what number, $y$, has $\log_b(y) = -1$? Think about what that actually means.

BTW: either use X or x, but not both in the same problem.

 

[Alykayy]

I figured them out, thank you.

log x - log (x+11) = -1
log (x / (x+11)) = -1
x/(x+11) = 10^-1
x/(x+11) = 0.1
x=0.1(x+11)
x=0.1x+1.1
x-0.1x=1.1
0.9x=1.1
x=1.222...

log₄x-log₄(x+15) =-1
log₄(x/(x+15) = -1
x/(x+15)= 4^-1
x/(x+15) = 0.25
x=0.25(x+15)
x=0.25x+3.75
x-0.25x=3.75
0.75x=3.75
x=5