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4 years ago

6

What is the solution to log5(10x-1)=log5(9x=7)

1 answer:

zepelin [54]4 years ago

4 0

$Domain:\\\\10x-1 > 0\ and\ 9x+7 > 0\\\\10x > 1\ and\ 9x > -7\\\\x > \dfrac{1}{10}\ and\ x > -\dfrac{7}{9}\Rightarrow x > \dfrac{1}{10}\\\\D=\left\{x|x > \dfrac{1}{10}\right\}\\\\\log_5(10x-1)=\log_5(9x+7)\iff10x-1=9x+7\qquad\text{add 1 to both sides}\\\\10x=9x+8\qquad\text{subtract 9x from both sides}\\\\x=8\in D\\\\Answer:\ \boxed{x=8}$

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professor190 [17]

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3 years ago

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Molodets [167]

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4 years ago

For an acute angle 0, the equation sin(0) = cos(6) is true. What is the value of ? Show or explain your answer

weqwewe [10]

Answer:

$\theta =84$

Step-by-step explanation:

Given

$sin(\theta) = cos(6)$

Required

Find $\thata$ $\theta$

In trigonometry:

If sin(A) = cos(B), then

$A + B = 90$

So, we have:

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Collect Like Terms

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$\theta = 90-6$

$\theta =84$

6 0

3 years ago

Please help me :) i would appreciate it a lot

Kobotan [32]

Answer:

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3 years ago

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OleMash [197]

Answer:

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3 years ago

Read 2 more answers