2 log2 2 + 2 log2 6 – log2 3x = 3 .. the base is 2?
I have the following mathematical expression:
2 log₂(2) + 2 log₂(6) - log₂(3x) = 3.
Is the base indeed 2? I would appreciate any clarification on how to solve this equation and its implications for x. Thank you!
4 Answers
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
the answer should be 6.
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
x = 6……ANSWER
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