How do you solve the logarithmic equation 3^2x=7^(x-1)?

Answer 1

take the log on both sides

and note that log (a^x) = xlog(a)

Answer 2

log[base 2](x + a million) + log[base 2](x - a million) = 3 First, use the log identity to integrate the sum of logs into the log of a product. log[base 2] [ (x + a million)(x - a million) ] = 3 Convert to exponential form, 2^3 = (x + a million)(x - a million) strengthen and simplify, 8 = x^2 - a million 9 = x^2 consequently, our ability recommendations are x = {-3, 3}. With logarithmic equations, you ought to ascertain for extraneous recommendations by ability of attempting out each and each fee into the unique equation. enable x = -3: Then LHS = log[base 2](x + a million) + log[base 2](x - a million) = log[base 2](-2) + .... OOPS! we can not take the log of a adverse quantity. Reject x = -3. If x = 3, we ought to continually be ok. consequently, our purely answer is x = 3.

Answer 3

Hi,
Well, this is one of those things that you can take the log of:
3^(2x) = 7^(x-1)
ln (3^(2x)) = ln (y^(x-1))
2x*ln3 = (x-1)* ln7 (Propetty of logs)
2x = (x-1) ln7/ln3
2x = (x-1)*1.771243 (Be sure you do ln7/ln3, not ln(7/3.)
2x -1.771242x = -1.771243
.228758x = -1.771243
x = -7.742868
or roughly -7.74

Hope this helps.
FE

Answer 4

First, take the log of both sides
2x*ln(3)=(x-1)*ln(7) then just reduce it as you would any other equation (since the ln terms are constants)

HTH

Doug

Answer 5

I post a question on Y!

Answer 6

3^2x=7^(x-1)--------take Log
Log3^2x=Log7^(x-1)
2xLog3=(x-1)Log7
2xLog3=xLog7-1Log7
2xLog3-xLog7=-log7
x(2Log3-log7)=-Log7
x=-Log7/(2Log3-Log7)=-7.743

Answer 7

by studying

Answer 8

I would ask a math teacher................

Or the poster below me

Answer 9

(2X)(log3) = (x - 1)(log7)

etc.