1/log(10)a + 1/log(10)a = 1/log(100)a?

Question

1/log(10)a + 1/log(10)a = 1/log(100)a How do I prove this? not solve for a, I need to prove it.

Answer 1

Hey! I have answered this before.
Assume that Log(a) [b] denotes the log of b to the base a.
Now Log(a) [b] = Log(x) [b]/ Log(x) [a]
So 1/log(10) [a] = = 1/ (log[a] / log[10]) = Log [10]/Log[a]=Log(a) [10]

So the sum reduces to Log (a)[10] + Log(a)[10]

Now log [a] + log[b] = log[ab]
So we get log(a)[10] + log(a)[10] = Log(a)[10 * 10] = Log(a)[100]
now log(a) [100] = Log[100] / Log[a] = 1/ (log[a]/Log[100]
= 1/Log(100) [a]

Answer 2

log(10)a means log of "a" base 10.
log(100)a means log of "a" base 100

log(100)a = log(10)a/log(10)100 = 0.5* log(10)a

so now that we have the RHS expressed as base 10 logs, lets just use L as the symbol for log.

so now we need to show that
1/La + 1/La = 1/(0.5*La)

we can see that the RHS simplifies

1/La + 1/La = 2/La

which is true for all values of a.

all done. of course a needs to be non zero and positive....

I made us of the conversion of logs from one base to another:

Answer 3

1/log(10)a+1/log(10)a
=2/log(10)a
=2log(a)a / log(10)a
=2log(a)10 / log (a) a
=2log(a)10
=log(a)100
=1/log(100)a

Answer 4

1/ loga base 10 + 1/loga base 10

log10/ log a + log10 / log a

log10 + log 10 / loga ,

log (ab) = log a x log b

log100/loga

1/log1oobase a

Answer 5

1/(log(10)a) + 1/(log(10)a) = 1/(log(100)a)
(1 + 1)/(log(10)a) = 1/(log(100)a)
2/(log(10)a) = 1/(log(100)a)
2/((log(a))/(log(10))) = 1/((log(a))/(log(100)))
2/(log(a) = 1/(log(a)/2)

cross multiply
2(log(a)/2) = log(a)

log(a) = log(a)

as you can see both sides are equal.