# 2 log2 2 + 2 log2 6 – log2 3x = 3 .. the base is 2?

• log(base2)[2² * 6² / 3x] = 3

144 / 3x = 2^3 = 8

144/8 = 3x

18 = 3x

x = 6

• 2 log2 (2) + 2 log2 (6) – log2 (3x) = 3

step 1) use the power formula of logarithms: x logb (a) = logb a^x: so, 2 log2 (2) + 2 log2 (6) – log2 (3x) = 3 becomes log2(2^2) + log2 (6^2) – log2 (3x) = 3

step 2) simplify: log2 (4) + log2 (36) – log2 (3x) = 3

step 3) use the product rule of logarithms: logb (a) + logb (c) = logb (ac). So, log2 (4) + log2 (36) – log2 (3x) = 3 becomes log2 (36*4) – log2 (3x) =3

step 4) simplify: log2 (144) – log2 (3x) = 3

step 5) use the quotient rule of logarithms: logb (a) – logb (c) = logb (a/c). So, log2 (144) – log2 (3x) = 3 becomes log2 (144/3x) = 3

step 6) transform the logarithmic equation into an exponential one: logb (x) = y is the same as b^y=x

So, log2 (144/3x) = 3 becomes 2^3 = 144/3x

step 7) Simplify: 8 = 144/3x

step 8) multiply both sides by 3x: 3x*8 = (144/3x)*3x

step 9) simplify: 24x = 144

step 10) divide both sides by 24 and simplify: 24x/24 = 144/24

• 2log2 2=log2 2^2=log2 4

2log2 6=log2 36

log2[(144/3x)]=3

log2(48/x)=3

48/x=8

x=6.

• LOG(2) 2^2 + LOG(2) 6^2 – LOG(2) (3X) = 3

Use log A +log B = log A*B AND logA – logB = Log A / B

log(2) 4 + log(2) 36 – log(2) (3x) = 3

……….4*36

log(2) ——….=…3

………..3x

………..144

log(2) ——….=…3

………..3x

………..48

log(2) ——….=…3

…………x

NOW Remember……log(B) A = C same as A = B^C

48

—…=…2^3

x

48 = 2^3 * x

48 = 8x