How do you solve #log_10 root3(10)=x#?

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Answer 1 · 2018-07-09T07:43:50

#x=1/3#

Log form and index form are interchangeable.

#log_a b = c " "hArr" " a^c =b#

Apply that to the equation given:

#log_10 root3(10) =x" "hArr" " 10^x = root3(10)#

To solve an exponential equation, one approach is to make the bases the same:

#10^x= 10^(1/3)#

#:. x =1/3#

We could also have raised the whole equation to the power of #3#

#(10^x)^3 = (root3(10))^3#

#10^(3x) = 10^1#

#3x =1#

#x =1/3#

Answer 2 · 2018-07-09T08:20:47

#x=1/3=0.bar3#

Given: #log_10root(3)(10)=x#.

A basic property of logs is that:

If #log_ab=m# then #a^m=b#.

So, we get:

#10^x=root(3)10#

Note that #root(3)10=10^(1/3)# due to the fact that #root(a)(m^n)=m^(n/a)#.

Therefore,

#10^x=10^(1/3)#

It is now clear to see that #x=1/3#. In decimal form, that is #0.3333...=0.bar3#.