5^x=20 log 5^x=log20 x log5 =log(5×2^2) x=(log5+log2^2)/ /log5 x=1+2 log5(2)
@MasterTMathsClass45 минут бұрын
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@AlexanderSemashkevichКүн бұрын
x=log20/log5=(log5+2log2)/log5=1+2log2/log5 x≈1+(2×0.301)/0.699≈1301/699≈1.861 Deviation is about 0.0001 3×5^(1301/699)≈59.99
@samuelkelly2197Күн бұрын
Samash, what do you intent to show by your solution and the deviation of 0.0001. Having even a deviation of 0.00000001 doesn’t equate to 60 = 60. Please do come again, thanks