How to solve log2(4)^cosx+log100^sinx=log3(1)?

OK, here you go sweetheart :)


I think this is what you mean :

log₂[(4)^(cosx)] + log₁₀[100^(sinx)] = log₃(1)

Using 'log laws' we can reduce the equation to :

log[4^(cosx)] /log2 + log[100^(sinx)] /log10 = log1 /log3

cosx (log4/log2) + sinx (log100/log10) = 0

2cosx + 2sinx = 0

cosx + sinx = 0


We now have a trigonometric equation left to solve!

Yaaaaaay !!!!!!!!!!!!!!!!!!!!!

Let's solve it, shall we?!

cosx + sinx = 0
sinx = - cosx
sinx/cosx = - cosx/cosx
tanx = - 1
x = - π/4


Since you haven't given us a domain to work with sweetheart, I'm going to assume you want all x ∈ R :)



General solution :

x = πn - (π/4) = π[n - (1/4)], where 'n' ∈ Z



Hope this helps !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

i) It is considered given, log2(4)^cosx as {log₂4}^cos(x)
= cos(x)*log₂4 [Applying power law of log]
Also, log₂4 = log₂(2^2) = 2*log₂2 [Applying power law of log] = 2 [As log₂2 = 1]
Thus {log₂4}^cos(x) = 2*cos(x)

ii) log(100) = 2
==> (log 100^sin(x) = 2*sin(x)
Also log₃1 = 0 [log 1 to any defined base = 0]

iii) Thus given equation reduces to: 2*cos(x) + 2*sin(x) = 0
==> cos(x) + sin(x) = 0

Dividing by cos(x) both sides and rearranging, tan(x) = -1
So x = 3π/4

In general, x = nπ - (π/4), where n is any integer.

We can solve this question by telling your teacher that you're cheating the homework on Yahoo answers, use your own dead brain!