What has been done here in this step?

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Comments (5)

A
anonymousfr
•11mos

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x=elog⁡5yx = e^{\log_5 y}

Using change of base:

log⁡5y=ln⁡yln⁡5\log_5 y = \frac{\ln y}{\ln 5}

Therefore:

x=eln⁡yln⁡5x = e^{\frac{\ln y}{\ln 5}}

Rewrite the exponent:

x=(eln⁡y)1ln⁡5x = \left(e^{\ln y}\right)^{\frac{1}{\ln 5}}

Since:

eln⁡y=ye^{\ln y} = y

So:

x=y1ln⁡5x = y^{\frac{1}{\ln 5}}

Now:

1ln⁡5=log⁡5e\frac{1}{\ln 5} = \log_5 e

Thus finally:

x=ylog⁡5ex = y^{\log_5 e}

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Change of base identity is used inside the exponents!

It is known as exponential-logarithm identity

The identity is

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X
xyzzzzzzz
OP•11mos

if i am not wrong isn't it when a=b, we can interchange a and c

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The identity is you can change base inside the exponents!

In most of the cases, it is used when a=b because it gives 1 in the exponent and result becomes the initial base itself

(Image: this happens in the power)

But that doesn't specifically mean ki voh bss equal ho tbhi use kr skte h!

I hope it is understandable!

Feel free to ask if you still have any doubt ;!

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