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devlian [24]
3 years ago
10

What is the true solution to 2ln(e ^ ln(5x)) - 2ln(15) of

Mathematics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

First you should see that the 2s cancel out. The equation becomes:

ln( e^{ln(5x)} ) = ln15ln(e

ln(5x)

)=ln15

Use the laws of logarithms, knowing that ln(e) = 1.

\begin{gathered}ln(5x) = ln(15) \\ 5x = 15 \\ x = 3\end{gathered}

ln(5x)=ln(15)

5x=15

x=3

The answer is B.

Hopefully this helps!

Step-by-step explanation:

x=3

Step-by-step explanation:

Given in an equation as

2 ln e^ln5x=2 ln 152lne

l

n5x=2ln15

Divide 2 to get

ln e^ln5x= ln 15lne

l

n5x=ln15

Using log rules for exponents we get

ln 5x = ln 15

Cancel ln to get

5x=15x=35x=15x=3

Hence solutin is x=3

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