1) Log(2)8^x= - 3 2) 5e^0.2x=7 The 2 in parenthesis is the base. What's the value of x?

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1) Log(2)8^x= - 3 2) 5e^0.2x=7 The 2 in parenthesis is the base. What's the value of x?

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Answer 1 · 2015-10-10T19:26:33

I found:
$x_{1} = - 1$ $x_1=-1$
$x_{2} = 1.6824$ $x_2=1.6824$

Explanation:

1) you can write (definition of log):
${log}_{2} (8^{x}) = - 3$ $log_2(8^x)=-3$
$8^{x} = 2^{- 3}$ $8^x=2^-3$
$8^{x} = \frac{1}{2^{3}} = \frac{1}{8}$ $8^x=1/2^3=1/8$
$8^{x} = 8^{- 1}$ $8^x=8^-1$
so taking the ${log}_{8}$ $log_8$ on both sides;
$x = - 1$ $x=-1$

2) $5 e^{0.2 x} = 7$ $5e^(0.2x)=7$
$e^{0.2 x} = \frac{7}{5}$ $e^(0.2x)=7/5$ taking the $ln$ $ln$ on both sides:
$0.2 x = ln (\frac{7}{5})$ $0.2x=ln(7/5)$
$x = 1.6824$ $x=1.6824$

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1) Log(2)8^x= - 3 2) 5e^0.2x=7 The 2 in parenthesis is the base. What's the value of x?

1 Answer

Explanation:

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