Solve 6^(2x + 4) = 2^(8 + x) * 3^(3x) for x

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Math wiz ya rite 09

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Could someone please help me solve this math question?

Note: x is a real number.

6^(2x + 4) = 2^(8 + x) * 3^(3x)

THANKS!
 
6^(2x+4)=2^(8+x)*3^3x take log
(2x+4) log 6 = log[2^(8+x)] + log [3^(3x)
(2x+4) log 6 = (8+x)log2 +3x log 3
2xlog6 +4 log 6 =8 log 2 +x log 2 +3x log 3
2x log 6 -x log2 -3xlog3=8 log 2 -4 log 6
x[2log6-log2-3log3]=[8log2-4log6]
x= [8log2-4log6] / [2log6-log2-3log3] answer

I do not have log tables available so will leave it in this form
Arthur
 
Math wiz ya rite 09 said:
Could someone please help me solve this math question?

Note: x is a real number.

6^(2x + 4) = 2^(8 + x) * 3^(3x)

THANKS!
Click to expand...

6 = 2*3
So,
6<SUP>2x + 4</SUP> = (2*3)<SUP>2x + 4</SUP>

6<SUP>2x + 4</SUP> = 2<SUP>2x + 4</SUP>*3<SUP>2x + 4</SUP>

Now, you can write your equation as

2<SUP>2x + 4</SUP>*3<SUP>2x + 4</SUP> = 2<SUP>8 + x</SUP>*3<SUP>3x</SUP>

Divide both sides of the equation by 2<SUP>8 + x</SUP>

Remember that when you divide powers of the same base, you SUBTRACT the exponents:

2<SUP>(2x + 4) - (8 + x)</SUP>*3<SUP>2x + 4</SUP> = 3<SUP>3x</SUP>

or,
2<SUP>x - 4</SUP>*3<SUP>2x + 4</SUP> = 3<SUP>3x</SUP>

Next, divide both sides by 3<SUP>2x + 4</SUP>, and remember again that when you divide powers of the same base, you subtract the exponents:

2<SUP>x - 4</SUP> = 3<SUP> 3x - (2x + 4)</SUP>

2<SUP>x - 4</SUP> = 3<SUP>x - 4</SUP>

Now, the only powers of 2 and 3 that are equal are 2<SUP>0</SUP> and 3<SUP>0</SUP>.....both of those are equal to 1.

So, x - 4 = 0
x = 4
 
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