# I need to know how to solve: `-9 log_11 2x =0`

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To solve this, you actually look at this like a normal equation, which makes it pretty easy! Let's look:

`-9log_11(2x) = 0`

It looks like we can start by dividing out that -9, which will just get rid of it!

`log_11(2x) = 0`

Now, we make both sides a power of 11. Remember that taking a value to an exponent is the inverse of the logarithm function:

`11^(log_11(2x)) = 11^0`

Now, because 11^ and `log_11` are inverse functions, the basically cancel out like here:

`2x = 11^0`

Now, remember that any number taken to the power of zero is 1:

`2x = 1`

Now we can solve finally by dividing by 2:

`x = 1/2`

And there's your answer!

Another way to think of this without algebraic solutions is to recognize that every log is 0 if and only iff the term inside the log is 1. The base doesn't matter (because EVERYTHING taken to the 0 power is 1). So, if you recognize this, you could start off the problem by saying "Well, the equation already tells me that 2x = 1."

And you may have been able to solve from there. But either way works great! I hope that helps!

**Sources:**

You need to use power rule of logarithms, hence `-9log_11(2x) = log_11(2x)^(-9) = log_11 (1/((2x)^9))`

Writing the new form of equation yields:

`log_11 (1/((2x)^9)) = 0 =gt (1/((2x)^9)) = 11^0 `

`(1/((2x)^9)) = 1 =gt ((2x)^9) = 1 =gt x^9 = 1/2^9`

`x = root(9)(1/2^9) =gt x = 1/2`

**Hence, the solution to the equation is `x = 1/2` .**