Solve the equation Log2(x2+5x-2)=log2(x2+3x-6)+log49

Question

this question is on engineering mathematics

Answer 1

log[2(x²+5x-2)]=log[2(x²+3x-6)]+log(49)=log[98(x²+3x-6)],

2(x²+5x-2)=98(x²+3x-6),

x²+5x-2=49(x²+3x-6)=49x²+147x-294,

48x²+142x-292=0,

24x²+71x-146=0,

x=(-71±√(5041+4×24×146))/48=(-71±√19057)/48=1.3968 or -4.3551 approx.

When x=1.3968, x²+5x-2>0 and x²+3x-6>0 so this solution is valid because the logs can be computed.

When x=-4.3551, x²+5x-2<0 and x²+3x-6<0 so this solution is invalid because the logs cannot be computed.

SOLUTION x=1.3968 approx.