Question

How do you solve `log (x+9) - log x = 1`?

Answer 1

`x = 1`

Explanation:

1) Determine when the equation is defined

First of all, let's determine for which `x` your equation is defined and for which `x` it is not defined.

Any logarithmic expression is only defined if its argument is greater than `0`.

So, in your case, `x + 9 > 0 <=> x > -9` and `x > 0` must hold.

As `x > 0` is the more restrictive condition of the two, the domain of your function is `x > 0`.

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2) Simplify the logarithmic equation

As next, you should combine all your logarithmic expressions into one.

You can do this with the logarithmic law

`log_a(n) - log_a(m) = log_a(n)/log_a(m)`

Thus, you equation can be transformed as follows:

`log(x+9) - log x = 1`

`<=> log((x+9)/x) = 1`

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3) Eliminate the logarithmic term

Now, as you haven't specified the base of the logarithm, I will assume that the base is `10`.

The inverse function of `log_10(x)` is `10^x` which means that both
`log_10(10^x) = x` and `10^(log_10(x)) = x` hold.

Thus, to eliminate the logarithmic expression at the left side, you need to apply the function `10^x` to the both sides of the equation:

`<=> 10^(log_10((x+9)/x)) = 10^1`

`<=> (x+9)/x = 10`

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4) Solve the equation

`(x+9)/x = 10`

... multiply both sides with `x` ...

`<=> x + 9 = 10x`

`<=> 9 = 9x`

`<=> x = 1`

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5) Check if the result is valid

We need to check if our result `x = 1` is consistent with the condition that we have computed in the beginning, `x > 0`.

Here, this is the case, thus we can accept `x = 1` as the solution of the equation.