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