How do you solve #-\log 8= 3- \log x#?

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Answer 1 · 2017-05-13T02:31:12

Isolate #logx# then use the definition of a logarithm to solve for #x#. Answer: #8000#

Question: #-log8=3-logx#

Note that #log(x)# is the same as #log_(10)(x)#

We can see that #-log8# and #3# are both constants, so we move them to one side and make #logx# positive:
#logx=3+log8#

Now we can use the definition of a logarithm (#log_(b)a=c -> b^c=a#) to take #x# out of the log:
#x=10^(3+log8)#

#=10^3*10^(log8)#

Since, exponential and logarithmic are inverse functions, they cancel each other:
#=1000*8#

#=8000#