If \(\displaystyle log_{10}A = log_{10}B\) then \(\displaystyle A=B\). Does that help?how to solve log 10 to the power of log (A-)/(HA) = log10 to the power of 4? Thanks
Hi, thanks a lot. You're right. The text has some printing error , so it's log_{10} (x/y). It's solvedHi, thanks a lot. But then to be exact, it is not log 10 ( x / y ), it is more of log ( x/y). Does it in this does not make sense?