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docker41 [41]
3 years ago
11

4^x-3=6 please help

Mathematics
1 answer:
SashulF [63]3 years ago
3 0
4^{x+3}=6\ \ \ \ |\log_4\\\\\log_44^{x+3}=\log_46\ \ \ |\text{Use}\ \log_ab^n=n\log_ab\\\\(x+3)\log_44=\log_46\ \ \ \ |\text{Use}\ \log_aa=1\\\\x+3=\log_46\ \ \ |-3\\\\x=\log_46-3\ \ \ \ |\text{Use}\ b=\log_aa^b\\\\x=\log_46-\log_44^3\\\\x=\log_46-\log_464\ \ \ |\text{Use}\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\x=\log_4\dfrac{6}{64}\\\\x=\log_4\dfrac{3}{32}
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Answer:

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Step-by-step explanation:

The amount at which Dan buys the car, PV = £2200

The rate at which the car depreciates, r = -0.5%

The car's worth, 'FV', in 6 years is given as follows;

FV = PV \cdot \left ( 1 + \dfrac{r}{100} \right )^n

Where;

r = The depreciation rate (negative) = -0.5%

FV = The future value of the asset

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n = The number of years (depreciating) = 6

By plugging in the values, we get;

FV = 2200 \times \left ( 1 + \dfrac{-0.5}{100} \right )^6 \approx 2,134.82

The amount the car will be worth which is its future value, FV after 6 years is FV ≈ £2,134.82 (after rounding to the nearest penny (hundredth))

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