newbie: solving variable in exponent of a scientific notated term

[Mike]

Hi All,

I can't seem to solve the following equation on paper without the help of
Matlab ... but I want to understand where I am going wrong here when
attempting to solve for Y:

359,259 = 19.97 x 10^ (0.1977(Y-1960))

My approach is to take the log of both sides:

log(359,259) = log ( 19.97 x 10^ (0.1977(Y-1960)) )

log(359,259) = (0.1977(Y-1960)) * log (19.97x10)

(0.1977(Y-1960)) = log(359,259) / log (19.97x10)

Y-1960 = log(359,259) / ( log (19.97x10) * 0.1977)

Y = log(359,259) / ( log (19.97x10) * 0.1977) + 1960

Y = 1972 which is incorrect. The answer should be 1981.5.

I suspect when I take the log of term in scientific notation which has Y in
the exponent. Can someone please clarify how you properly take the log of
scientific notated term? I did find that if I set the right hand log to:
log(19.97), I got approx the right answer.

However, when I applied the same "logic" to another similar equation:

65705 = 1027 x 10^(0.1505(Y-1970))

it didn't work so I suspect, there's more too it. When I do the same with
an exponential term: e^(x+5) then my resulting log term is: (x+5)*log(e)
and this is correct so it really must have something to do with the way I am
taking of the log of a scientifically notated term.

Thanks in advance, Mike.

[Christopher Creutzig]

Mike wrote:

> log(359,259) = log ( 19.97 x 10^ (0.1977(Y-1960)) )
>
> log(359,259) = (0.1977(Y-1960)) * log (19.97x10)

No, check your parentheses and operator precedences.

log(19.97 x 10^(0.1977*(Y-1960))
= log(19.97) + 0.1977*(Y-1960)*log(10)

Regards,
Christopher

[Jean-Marc Gulliet]

Mike,

Your approach is correct. However, you seem to have forgotten the
product rule for logarithms: if a and b are two reals such that a*b>0,
then log(a*b) == log(a) + log(b). In other words, the logarithm of a
product is equal to the sum of the logarithms of the terms.

HTH,
Jean-Marc

[Christopher Creutzig]

Jean-Marc Gulliet wrote:

> Your approach is correct. However, you seem to have forgotten the
> product rule for logarithms: if a and b are two reals such that a*b>0,
> then log(a*b) == log(a) + log(b). In other words, the logarithm of a

Minor correction: The condition should read “if a > 0 or b > 0.”
0=log(1)=log(-1*-1)=/=log(-1)+log(-1)=2*I*PI. (I regard log as a
single-valued, well-defined function on C\{0}. If you don't, the
equation does not need a restriction.)

Regards,
Christopher

[Christopher Creutzig]

Christopher Creutzig wrote:

> Minor correction: The condition should read “if a > 0 or b > 0.”

Or, to give a condition that is both necessary and sufficient:
-PI<arg(a)+arg(b)<=PI.

Regards,
Christopher

[Mike]

Thank you everyone !

"Christopher Creutzig" <chris...@creutzig.de> wrote in message
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