finding the inverse of h(x) = [5 * 7^(2x - 1) + 3] / 4

[Math wiz ya rite 09]

Find the inverse of the function:

h(x) = [5 * 7^(2x-1) + 3] / 4

Thanks for your time!

 

[stapel]

Follow the usual procedure:

. . . . .i) Rename "h(x)" as "y".
. . . . .ii) Solve for "x=".
. . . . .iii) Switch "x" and "y".
. . . . .iv) Rename the new "y" as "h^-1(x)".

It looks like you'll need to use logs during step (ii).

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.

 

[Math wiz ya rite 09]

y = [5 * 7^(2x-1) + 3] / 4


i can do step 1.. its a no brainer, but i don't know how to use logs in the second step???

please help?

 

[Math wiz ya rite 09]

y = [5 * 7^(2x-1) + 3] / 4

4y =[5 * 7^(2x-1) + 3]

4y - 3 = [5 * 7^(2x-1)]

(4y - 3) / 5 = 7^(2x-1)

 

[stapel]

i don't know how to use logs....

So logarithms have not been covered yet in class? In that case, I'm afraid I don't see how you're supposed to work this. Sorry.

Eliz.

 

[Math wiz ya rite 09]

y = [5 * 7^(2x-1) + 3] / 4

4y =[5 * 7^(2x-1) + 3]

4y - 3 = [5 * 7^(2x-1)]

(4y - 3) / 5 = 7^(2x-1)

would it be log base 7 of (4y - 3) / 5 = 2x-1

???

 

[stapel]

y = [5 * 7^(2x-1) + 3] / 4

4y =[5 * 7^(2x-1) + 3]

4y - 3 = [5 * 7^(2x-1)]

(4y - 3) / 5 = 7^(2x-1)

would it be log base 7 of (4y - 3) / 5 = 2x-1

Oh, so you are familiar with logarithms. I'm sorry; I must have misunderstood your previous reply.

Yes, taking logs of both sides would be done in this manner. The particular base doesn't matter, though base-10 or base-e would be easier for exercises in which you need an approximate decimal answer. (Of course, you can use the change-of-base formula to convert base-7 answers to either of these, if/when necessary.)

Now solve for "x=", switch the variables, and do the second renaming. And you're done!

Eliz.

 

[Math wiz ya rite 09]

log base 7 of (4y - 3) / 5 = 2x-1



[log base 7 of (x - 3) / 10] + 1 = y

i don't need to get a decimal approximation, so would my answer be correct. Is that the most simplified it gets?

 

[Math wiz ya rite 09]

or should i add ten to the numerator cause 10/10 is 1, so it would seem as a way to simplify??

 

[stapel]

How did you go from:

. . . . .log₇[(4y - 3) / 5] = 2x - 1

...to:

. . . . .log₇[(4y - 3) / 10] + 1 = x

I understand the "plus one", but what happened to the "2", and where did the "10" come from?

Thank you.

Eliz.

 

[Math wiz ya rite 09]

to get x by itself, i divided by 2. so 5 was the denominator divide by 2 to get 10 rite?

 

[stapel]

The "5" was inside the logarithm. Dividing the log by 2 just gives you "(log[whatever]) / 2". The "log" is a function, not a variable; you can't treat it like multiplication or something.

Eliz.