Question #39b7b

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Question #39b7b

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Answer 1 · 2017-12-15T13:44:56

No solution.

Explanation:

$7^{ln x} = e^{λ ln x} = {(e^{ln x})}^{λ} = x^{λ}$ $7^lnx = e^(lambda lnx)=(e^lnx)^lambda = x^lambda$

but $λ = ln 7$ $lambda = ln7$ so finally we have

$x^{ln 7} - x^{ln 7} = 0 \neq 686$ $x^ln7-x^ln7=0 ne 686$ and the equations has not solution.

Answer 2 · 2017-12-15T13:47:04

There is no solution. Had the sign between the two terms been $+$ $+$ instead of $-$ $-$ , then there would be a solution. Check the problem.

Explanation:

We have $ln (a^{b}) = b ln a$ $\ln(a^b)=b\ln a$ , so:

$ln (7^{ln x}) = (ln x) (ln 7)$ $\ln(7^{\ln x})=(\ln x)(\ln 7)$

$ln (x^{ln 7}) = (ln 7) (ln x)$ $\ln(x^{\ln 7})=(\ln 7)(\ln x)$

The logarithms are equal, therefore the numbers $7^{ln x}$ $7^{\ln x}$ and $x^{ln 7}$ $x^{\ln 7}$ are equal, and their difference is never anything other than zero.

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Question #39b7b

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