Identities Question?

Question

I don't know how to prove 1/Log a + 1/Log a = 1/log 100 a (10 is the base)

This is an identity question

Answer 1

1/Log(10) [a] = Log(a) [10]
() indicates the base

Log(a) [10] + Log(a) [10] = Log(a) [10*10] = Log(a) [100]

This is the same as 1/Log(100) [a]

Answer 2

This is an equation , you have to find a not to prove it : 1/Log a + 1/Log a = 1/log 100 a ====>
log a + log a = log 100 a
2 log a = log 100 a
log a^2 = log 100 a ==> a^2 = 100 a ====>
a = 10 radical a

Answer 3

1/log a + 1/log a = 2/log a = log 100/log a = 1/(log a/log 100) = 1/log_100 a.

Answer 4

Consider log_M*N = log_M + log_N

Let Log_M = s
and Log_N = t
The equivalent forms are;
M = x^s
N = x^t
M*N = x^s * x^t = x^s+t
Therefore;
log_M*N = s+t
Therefore;
log_M*N = log_M + log_N
1/(log_M*N) = 1(log_M + log_N)
1/(log_M*N) = 1/(log_M) + 1/(log_N)