# Log questions

## Homework Statement

Solve for X

logx-log(x+11) = -1

and

log4x-log4(x+15) = -1

## 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) = -1

I don't know how to get the X out of the log to solve for X

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Mark44
Mentor

## Homework Statement

Solve for X

logx-log(x+11) = -1

and

log4x-log4(x+15) = -1

## 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) = -1

I 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:
loga(x) = y is equivalent to x = ay.

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

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Solve for X

logx-log(x+11) = -1

and

log4x-log4(x+15) = -1

## 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) = -1

I 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.

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