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3 years ago

Find X : log2+ logx=1

Mathematics

2 answers:

Dominik [7]3 years ago

8 0

$log2+logx=1;\ D:x\in\mathbb{R^+}\\\\log(2\ \cdot\ x)=log10\iff2x=10\ \ \ /:2\\\\x=5\in D\\\\Solution:x=5.$

Vinil7 [7]3 years ago

3 0

$\log2+\log x=1\\
D:x>0\\
\log2x=1\\
10^1=2x\\
2x=10\\
x=5$

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Anna007 [38]

If you would like to solve 1/3 / 6, you can do this using the following steps:

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The correct result would be 1/18.

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3 years ago

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4 years ago

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r-ruslan [8.4K]

ANSWER:

The solution of the two equations 2x+3y=5 and 4x - y=17 is (4, -1).

SOLUTION:

Given, two linear equations are 2x + 3y = 5 → (1) and 4x – y = 17 → (2).

Let us first solve the above equations using elimination process. 

For elimination, one of the coefficients of variables has to be same in order to cancel them.

Now solve (1) and (2)

eqn (1) $\times$ 2 → 4x + 6y = 10

eqn (2) → 4x – y = 17

(-) ----------------------------

0x + 7y = -7

y = -1

Substitute y value in (2)

$4 x-(-1)=17 \rightarrow 4 x+1=17 \rightarrow 4 x=17-1 \rightarrow 4 x=16 \rightarrow x=4$

So, solution of two equations is (4, -1).

Now let us solve using substitution process.

Then, (2) → 4x – y = 17 → 4x = 17 + y → y = 4x – 17

Now substitute y value in (1) → 2x + 3(4x – 17) = 5 → 2x + 12x – 51 = 5 → 14x = 5 + 51 → 14x = 56

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Hence, the solution of the two equations is (4, -1).

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3 years ago