Solve log base tanx 3 + log base cotx 27 + 4 = 0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Supposing that you need to solve the equation `log_(tan x)3 + log_(cot x) 27 + 4 = 0` , you should substitute `3^3`  for `27`  such that:

`log_(tan x)3 + log_(cot x) (3^3) + 4 = 0`

Using the logarithmic identity `log_a b^n = n*log_a b`  yields:

`log_(tan x)3 + 3log_(cot x) 3 + 4 ` = 0

Using the logarithmic identity `log_a b = 1/log_b a`  yields:

`1/(log_3 (tan x)) + 3/(log_3 (cot x)) + 4 = 0`

You need to remember that `cot x = 1/tan x`  such that:

`1/(log_3 (tan x)) + 3/(log_3 (1/tan x)) + 4 = 0`

Using the logarithmic identity `log (a/b) = log a - log b`  yields:

`1/(log_3 (tan x)) + 3/(log_3 1- log_3 (tan x)) + 4 = 0`

You should remember that `log_3 1 = 0` ,such that:

`1/(log_3 (tan x)) - 3/(log_3 (tan x)) + 4 = 0`

`-2/(log_3 (tan x)) = -4 => 1/(log_3 (tan x)) = 2`

`2(log_3 (tan x)) = 1 => (log_3 (tan x)) = 1/2 => tan x = 3^(1/2)`

`tan x = sqrt 3 => x = pi/3 + npi`

Hence, evaluating the general solution to the given equation yields `x = pi/3 + npi.`

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