Show that: ( ln 10 ) ( log e ) = 1?

ln 10 = log 10 / log e

(ln 10 )(log e) = (log 10 / log e)*(log e) = log 10 = 1

change-of-base rule:

logᵤ(x) = logᵥ(x)/logᵥ(u)

ln(10) = logₑ(10)

= log₁₀(10)/log₁₀(e)

= 1/log₁₀(e)

ln(10)×log(e) = (1/log(e))×log(e) = 1

ln(10) log(e) ?

use the change of base formula on one factor

assume log is base 10

ln(10) = log(10)/log(e)

ln(10) log(e) = log(10) log(e)/log(e)

= log(10) = 1

ok.. from this relationship

.. loga(b) = logc(b) / logc(a)

where

.. loga(b) means log of b base a

.. logc(b) means log of b base c

.. logc(a) means log of a base c

and of course

.. ln means base "e"

.. log... without a base shown... means base 10... .log10

we can readily write

.. ln(10) = log(10) / log(e)

so that

..ln(10 * log(e) = [log10 / log(e)] * log(e) = log(10) = 1

ln is the natural log (base e)

log is the common log (base 10).

Using the base conversion formula:

ln 10 = log(10) / log(e) = 1 / log(e)

Then if you multiply that by log(e) you cancel the denominator.

1/log(e) * log(e) = 1

ln(10) log(e)

= log(10)/log(e) * log(e)/log(10) = 1

Put in the good bases.