It isn't hard, for you all, apart from me.
Xlog 3=(X+1)log 2
Please help!
Show me the steps...
2007-02-03 15:18:54 · 7 answers · asked by blackmail8549 2 in Science & Mathematics ➔ Mathematics
xlog3 = (x+1)log2
distribute
xlog3 = xlog2 + log 2
xlog3-xlog2 = log2
x(log3-log2) = log2
xlog(3/2) = log2
x= log2 / log(3/2)
x= 1.7095
2007-02-03 15:26:16 · answer #1 · answered by 7 · 1⤊ 0⤋
if you wrote this correctly, this is not a "log" equation, as the log3 and log2 are just constants (numbers). Just imagine how you would solve this if they had written 3 and 2 instead, and do the same thing:
xlog3 = xlog2 +1log2
xlog3 - x log2 = log 2
x(log3 - log2) = log 2
x = log2 / (log3 - log2)
2007-02-03 23:29:32 · answer #2 · answered by grand_nanny 5 · 0⤊ 0⤋
Xlog 3=(X+1)log 2
.4771*X=.3010*X+.3010
.1759*X=.3010
X=.3010/.1759=1.7095
2007-02-03 23:28:50 · answer #3 · answered by anonimous 6 · 0⤊ 0⤋
Remember that log 3 and log 2 are constants, so first treat them just like any other coefficients:
X*log(3)=(X+1) log(2)
X*log(3)=X*log(2) + log(2)
X*log(3) - X*log(2) = log(2)
X(log(3) - log(2)) = log(2)
x = log(2) / (log(3) - log(2))
x = log(2) / log(3/2)
2007-02-03 23:41:42 · answer #4 · answered by Anonymous · 0⤊ 0⤋
xlog(3) = xlog(2) + log(2
x(log(3) - log(2)) = log(2)
x = log(2)/(log(3) - log(2))
x = log(2)/(log(1.5)
x = 1.709511
2007-02-03 23:33:28 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋
xlog3 = xlog2 +1log2
xlog3 - x log2 = log 2
x(log3 - log2) = log 2
x = log2 / (log3 - log2)
2007-02-03 23:39:51 · answer #6 · answered by Anne-Ice♥cream 2 · 0⤊ 0⤋
first log3 and log2 are just numbers
xlog3=xlog2+log2
xlog3-xlog2=log2
x(log3-log2)=log2
x=log2/(log3-log2)
2007-02-03 23:31:55 · answer #7 · answered by n_ax1000 2 · 0⤊ 0⤋