SOLUTION: Could someone please help me with these problems? Solve the following equations. Round to 4 decimal places when necessary. a. Log3 x=8 b. 27^2x+2=81^x-3 c. log2 (3x+2)=3

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Could someone please help me with these problems? Solve the following equations. Round to 4 decimal places when necessary. a. Log3 x=8 b. 27^2x+2=81^x-3 c. log2 (3x+2)=3       Log On

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Question 60837: Could someone please help me with these problems?
Solve the following equations. Round to 4 decimal places when necessary.
a. Log3 x=8
b. 27^2x+2=81^x-3
c. log2 (3x+2)=3
d. e^3x=80
I am not sure how to make the numbers beside log small
I REALLY NEED HELP ON THESE!!!
Found 2 solutions by Earlsdon, uma:
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
You can always rewrite the logarithmic equation as an equivalent exponential equation:
logb%28M%29+=+N ---> b%5EN+=+M where b is the base of the logarithm.
a) log3%28x%29+=+8 ---> 3%5E8+=+x ---> 6561+=+x
b) 27%5E%28%282x%2B2%29%29+=+81%5E%28%28x-3%29%29 but 27+=+3%5E3 and 81+=+3%5E4, so:
%283%5E3%29%5E%28%282x%2B2%29%29+=+%283%5E4%29%5E%28%28x-3%29%29 Multiplying the exponents on both sides.
3%5E%28%286x%2B6%29%29+=+3%5E%28%284x-12%29%29 but, if x%5En+=+x%5Em then n+=+m...the exponents are equal.
6x%2B6+=+4x-12 Simplify and solve for x.
2x+=+-18
x+=+-9
c)log2%283x%2B2%29+=+3 Similar to problem a) Rewrite in exponential form.
2%5E3+=+3x%2B2
8+=+3x%2B2 Subtract 2 from both sides.
3x+=+6 Divide both sides by 3.
x+=+2
d) e%5E%283x%29+=+80 Take the natural log (because of the e) of both sides.
3xln%28e%29+=+ln%2880%29 Because ln%28e%29%5Ex+=+xln%28e%29. Recall that ln%28e%29+=+1
3x+=+4.3820
x+=+1.4607

Answer by uma(370) About Me  (Show Source):
You can put this solution on YOUR website!
1) Log 3 x = 8
==> log x to the base 3 = 8
Coverting this to exponential form, we have,
x = 3^8
x = 6561.

2) 27^(2x+2) = 81^(x-3)
Now 27 = 3^3 and 81 = 3^4
==> 3^3(2x+2) = 3^4(x-3)
==> 3^(6x+6) = 3^(4x-12)
As the bases are the same,
we equate the powers.
==> 6x+6 = 4x-12
==> 6x + 6 - 4x = 4x - 12 - 4x
==> 2x + 6 = - 12
==> 2x + 6 - 6 = -12 - 6
==> 2x = - 18
==> 2x/2 = -18/2
==> x = - 9


3) LOg 3x+2 to the base 2 = 3
In exponential form this becomes,
3x + 2 = 2^3
==> 3x + 2 = 8
==> 3x + 2 - 2 = 8 - 2
==> 3x = 6
==> 3x/3 = 6/3
==> x = 2

4)e^3x = 80
Taking logarithm on both the side we have..
log[e^3x] = Log 80
==> 3x = log 80 [AS e and log cancel each other]
==> 3x = 1.9031
==> 3x/3 = 1.9031/3
==> x = 0.6344

Good LUck!!!