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Serga [27]
3 years ago
11

3 log 2x=4 Please solve

Mathematics
1 answer:
Leno4ka [110]3 years ago
6 0

3 log 2x=4\ /:3\\log2x=\frac{4}{3}\\10^{\frac{4}{3}}=2x\ /:2\\x=\frac{10\sqrt[3]{10} }{2}\\x=5\sqrt[3]{10}

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2 years ago
How to solve using substitution<br> y=3/4X+5<br> 4x-3y=-1
Pachacha [2.7K]
Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as <span>y=(3/4)X+5

Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:

4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)

Let 's now solve the system   

4y=3x + 20
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We are to solve this system using subtraction.  To accomplish this, multiply the first equation by 3 and the second equation by 4.  Here's what happens:

12y = 9x + 60  (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)

Then we have 

 12y = 9x + 60
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If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x:  7x = 56; x=8

We were given equations   
</span><span>y=3/4X+5
4x-3y=-1

We can subst. x=8 into either of these eqn's to find y.  Let's try the first one:

y = (3/4)(8)+5 = 6+5=11

Then x=8 and y=11.

You should check this result.  Subst. x=8 and y=11 into the second given equation.  Is this equation now true?</span>
5 0
3 years ago
(-5,-5); y = - 5x + 5
Phoenix [80]

Answer:

Step-by-step explanation:

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3 years ago
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