Solve the exponential equation algebraically
2^x=33log2^x=log33xlog2=log33(xlog2)/log2 = (log33)/log2x=(log33)/log2= 5.04439411936
hmm, okso rememeber\(log_ab=c\) means \(a^c=b\)therefor\(a^c=b\) is to \(2^x=33\) as is\(log_ab=c\) is to \(log_233=x\)x=\(log_233\) or aprox 5.04439
