2 years ago

Solve 7e^(2x)-5>58. Round to the nearest hundredth.

Mathematics

1 answer:

7nadin3 [17]2 years ago

5 0

Answer:

$x\gtrapprox1.1$

Step-by-step explanation:

$7e^{2x}-5>58$

$7e^{2x}>63$

$e^{2x}>9$

$2x>ln(9)$

$x>\frac{ln(9)}{2}$

$x\gtrapprox1.1$

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