Question

Solve the equation `log_2 4^cosx+log100^sinx=log_3 1`?

Answer 1

`log_2(4)^cosx+log_10 (100)^sinx=log_3(1)`

`=>log_2(2)^(2cosx)+log_10 (10)^(2sinx)=0`

`=>2cosxlog_2(2)+2sinxlog_10 (10)=0`

`=>cosx+sinx=0`

`=>tanx=-1=tan(pi-pi/4)`
`=>x=(3pi)/4`

General solution

`x=npi-pi/4" where " n in ZZ`

Answer 2

`x=2npi+pi/4+-pi/2`, where `n` is an integer

Explanation:

Before we solve `log_2 4^cosx+log100^sinx=log_3 1`, a few facts

1. When we write `loga` it means the base is `10` (and when we write `lna` base is `e`).
2. As `a^0=1`, `log_a1=0` i.e. logarithm of `1` to any base `a` is `0`.

Hence `log_2 4^cosx+log100^sinx=log_3 1`

or `cosxlog_2 4+log(10^2)^sinx=0`

or `cosxlog_2 2^2+sinxlog10^2=0`

or `2cosx+2sinx=0`

or `cosx+sinx=0`

or `cosx*1/sqrt2+sinx*1/sqrt2=0`

or `cosxcos(pi/4)+sinxsin(pi/4)=0`

or `cos(x-pi/4)=0=cos(pi/2)`

Hence `x-pi/4=2npi+-pi/2`

and `x=2npi+pi/4+-pi/2`, where `n` is an integer