Hello,
The question is "Given log 2=0.3010 and log 3=0.4771, if 5^x-1=6^2-x,then find value of X".
sol: 5^x-1 = 6^2-x
log 5^x-1 = log 6^2-x
(x-1) log10/2=(2-x)(log(3*2))
(x-1)(log10-log2)=(2-x)(0.3010+0.4771)
(x-1)(0.6990)=(2-x)(0.7781)
o.990x-0.6990=1.5562-0.7781x
0.6990x+0.7781x=1.5562+0.6990
1.4771x=2.2552
x = 2.2552/1.4771
x=1.52689 (approx)
OR
5^x-1 = 6^2-x
log 5^x-1=log 6^2-x
(x-1) log10/2 =(2-x) log 3*2
(x-1) (log10-log2) = (2-x) (log2+ log 3)
(x-1) (1-0.3010) = (2-x) (0.3010 + 0.4771)
(x-1) (0.6990) = (2-x) (0.7781)
0.6990x-0.6990=1.5562-0.6990
x=2.2552/1.4771
x=1.5268 (approx)
NOTE:
(^ is caret or circumflex and it means to the power of)
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