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When solving for H3O+ with pH, how do i 'get' rid of log ('move it to the other side')
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forget about the log
H3O+ = 10^-pH
equation "pH=-log[H3O+)
alright thanks
this formula will always work H3O+ = 10^-pH
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\[x = b ^{y}\] y is the logarithm of x to base b, and is written \[y = \log _{b}(x)\] Or just use \[\log ^{-1}\] on your calculator.
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