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Varvara68 [4.7K]
3 years ago
12

Solve the logarithmic equation. Be sure to reject any value of log (X + 8) = log x + log 8​

Mathematics
2 answers:
iogann1982 [59]3 years ago
8 0

Answer:

x =  \frac{8}{7}

Step-by-step explanation:

log(x + 8)  =  log(x)  +  log(8)  \\ (x + 8) = 8x \\ x + 8 = 8x \\ 8 = 8x - x \\ 8 = 7x \\  \frac{8}{7} =  \frac{7x}{7}  \\ x =  \frac{8}{7}

swat323 years ago
6 0

Answer:

x = 8/7

Step-by-step explanation:

log (X + 8) = log x + log 8​

We know that log a + log b = log ab

log (X + 8) = log 8x

Raise each side to base 10

10^log (X + 8) = 10^log 8x

x+8 = 8x

Subtract x from each side​

x+8-x = 8x-x

8 = 7x

Divide by 7

8/7 = 7x/7

8/7 =x

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Consider the initial value problem:
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Answer:

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Step-by-step explanation:

The given initial value problem is;

y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2

Let

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Differentiating both sides of equation (1) with respect to t, we obtain:

u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)

Differentiating both sides of equation (2) with respect to t gives:

v'(t)=y''(t)----(6)

From equation (1),

y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)

Putting t=0 into equation (2) yields

u(0)=y(0)\\\implies u(0)=-5

Also putting t=0 into equation (3)

u'(0)=y'(0)\\u'(0)=2

The system of first order equations is:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

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