Inverse function of an exponential equation: 1/2(5)^(x+3)-1

[confused15]

So for the exponential equation y=1/2(5)^x+3-1 what would the inverse equation be? I know that x=1/2(5)^y+3-1 would be the inverse equation, but I am wondering how this would be written in logarithmic form. For example as in y=(5/7)^x and its inverse y=log_5/7 ​x . Thanks for the help ;D

 

[Steven G]

So for the exponential equation y=1/2(5)^x+3-1 what would the inverse equation be? I know that x=1/2(5)^y+3-1 would be the inverse equation, but I am wondering how this would be written in logarithmic form. For example as in y=(5/7)^x and its inverse y=log_5/7 ​x . Thanks for the help ;D

Solve \(\displaystyle x =\ \dfrac{1}{2}*5^{(y+3)}-1 \ \ for \ \ y\)

 

[stapel]

So for the exponential equation y=1/2(5)^x+3-1 what would the inverse equation be? I know that x=1/2(5)^y+3-1 would be the inverse equation, but I am wondering how this would be written in logarithmic form. For example as in y=(5/7)^x and its inverse y=log_5/7 ​x .

To find out, apply the usual inverse-finding process (here):

. . . . .\(\displaystyle y\, =\, \dfrac{1}{2\, (5)^{x+3}}\,-\, 1\)

. . . . .\(\displaystyle y\, +\, 1\, =\, \dfrac{1}{2\, (5)^{x+3}}\)

. . . . .\(\displaystyle \dfrac{1}{y\, +\, 1}\, =\, 2\, (5)^{x+3}\)

. . . . .\(\displaystyle \dfrac{1}{2\, (y\, +\, 1)}\, =\, 5^{x+3}\)

Now apply an exponent rule (here):

. . . . .\(\displaystyle 5^{x+3}\, =\, (5^x)\, (5^3)\, =\, 125\, (5^x)\)

Then:

. . . . .\(\displaystyle \dfrac{1}{2\, (y\, +\, 1)}\, =\, 125\, (5^x)\)

. . . . .\(\displaystyle \dfrac{1}{250\, (y\, +\, 1)}\, =\, 5^x\)

Then log both sides, etc, etc.

 

[Ishuda]

So for the exponential equation y=1/2(5)^x+3-1 what would the inverse equation be? I know that x=1/2(5)^y+3-1 would be the inverse equation, but I am wondering how this would be written in logarithmic form. For example as in y=(5/7)^x and its inverse y=log_5/7 ​x . Thanks for the help ;D

Same sort of thing as the rest:
\(\displaystyle x\, +\, 1\, =\, \frac{1}{2}\, 5^{y+3}\)
\(\displaystyle 2\, (x\, +\, 1)\, =\, 5^{y+3}\)
\(\displaystyle ln(2\, (x\, +\, 1))\, =\, ln(5^{y+3}\,\,)\)
\(\displaystyle ln(2\, (x\, +\, 1))\, =\, (y+3)\, ln(5)\)
\(\displaystyle (y+3)\, ln(5)\, =\, ln(2\, (x\, +\, 1))\)
\(\displaystyle (y+3)\, =\, \dfrac{ln(2\, (x\, +\, 1))}{ln(5)}\)
\(\displaystyle y\, =\, \dfrac{ln(2\, (x\, +\, 1))}{ln(5)}\, -\, 3\)
and carry on to
\(\displaystyle y\, =\, \dfrac{ln\large(\frac{2\, (x\, +\, 1)}{125}\, \large)}{ln(5)}\)