Question

2 Quick Algebra questions in preparation of a test. I am not really sure how to do these 2 types of problems, can someone explain to me how to solve them? Please show me all steps. :)

1). 6^3x=104
2). log3(x+2)+log3(4x-1)=5

Answer 1

1) 6^(3x)=104

loga =logarithm to the base a

loga n=z ⇔a^z=n

Therefore:
log6 104=3x
x=(log6 104) / 3
x=(ln 104) / 3ln 6=0.8640279....≈0.864 (loga z=ln a / ln z)

Answer: x≈0.864

2) log3 (x+2)+ log3 (4x-1)=5
log3 x=5 ⇔ 3⁵=x ⇒x=243
log3 [(x+2)(4x-1)]=log3 243
log3 (4x²+7x-2)=log3 243

Therefore:
4x²+7x-2=243
4x²+7x-245=0

we solve this square equation:
x=[-7⁺₋√(49+3920)] / 8 =(-7⁺₋63) / 8
we have two solutions:

x₁=(-7-63) / 8=-8.75 this solution is not valid.
x₂=(-7+63) /8=7

Answer: x=7