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butalik [34]
3 years ago
15

Condense the logarithm2 log c + 3 log k​

Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Given:

The expression is

2\log c+3\log k

To find:

The value after condense the logarithm.

Solution:

We have,

2\log c+3\log k

Using properties of logarithm, we get

=\log c^2+\log k^3            [\because \log x^n=n\log x]

=\log (c^2k^3)            [\because \log (ab)=\log a+\log b]

Therefore, the required value is \log (c^2k^3).

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For what values of N will the product 3n be less than 50
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3n<50 

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The variable N represents a whole number. 
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3 years ago
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(9x+2)(4x^2+35x-9)=0
pychu [463]
(9x+2)*(4x^2+35x-9)=0

we can make this...

9x+2=0

and

4x^2+35x-9=0

then...

9x+2=0

\boxed{x=-\frac{2}{9}}


4x^2+35x-9=0

x=\frac{-b\pm\sqrt{b^2-4*a*c}}{2*a}

x=\frac{-35\pm\sqrt{35^2-4*4*(-9)}}{2*4}

x=\frac{-35\pm\sqrt{1369}}{8}

x=\frac{-35\pm37}{8}

x_1=\frac{-35+37}{8}=\frac{1}{4}

x_2=\frac{-35-37}{8}=-9

\boxed{\boxed{S=\{\frac{1}{4},-\frac{2}{9},-9\}}}

I'm sure about my answer, don't matter the order...
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From my work, i got 4x+$1.50=48 I hope it helps you.

if i got it wrong please forgive me! also if it is Okie with you, Can i please have brainliest? Thank you!

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