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

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.