[Math] Solving Exponential Equations

[w00t22]

i was trying out some of my homework tonight and i stumbled across these questions

4^x+1 + 4^x = 160

2^x+2 + 2^x = 320

Answers

Spoiler

2.5

5

So first i thought same base (4) remove them, x + 1 + x = 160, ends up way to big. then i tryed the power law so

(x+1)log4 + xlog4 = 160
expand?
xlog4 + log4 + xlog4 = 160
simplified
x(log4+log4) = 160-log4 ends up like 132 which is wrong.

And another question for expanding the (x+1 and x-1) in front of logs, if its x+3, so
(x+3)log4
xlog4 + log12 or xlog4 + 3log4

this is my teachers first year teaching this course and hes not the best at lessons and helping students ><

thanks for any help plus rep of course

 

[Ghan]

No need for logs, really. But the answer is a bit elusive.

4^(x+1) + 4^x = 160
4^(x+1) = (4^x)(4^1)
(4)(4^x) + 4^x = 160
(5)(4^x) = 160
4^x = 32
(2^2)^x = 2^5
2^(2x) = 2^5
2x = 5
x = (5/2)

 

[w00t22]

i was confused at first since u wrote 2/5
x = (2/5)

but you meant 5/2 i think? maybe you write it different anyways = 2.5 thanks got to write it down to understand i hate the 2^x it confuses me

btw how did you get from
(4)(4^x) + 4^x = 160

to

(5)(4^x) = 160 5?

 

[phyrex1an]

And another question for expanding the (x+1 and x-1) in front of logs, if its x+3, so
(x+3)log4
xlog4 + log12 or xlog4 + 3log4

b*log(a) = log(a^b) so 3log(4) = log(64)

As for the reason why you got the wrong answer, you make two mistakes in your first step.
First you forgot to take the log of 160, right side should be log(160).
Second you used a kinda backwards version of one of the laws of logarithm.
log(4^x+1 + 4^x) isn't (x+1)log4 + xlog4. It was probably on this step your teacher figured you should give up and look for an alternative solution [at a forum]

 

[Ghan]

> but you meant 5/2 i think?

Yeah. Thanks for the catch. Hope you got the idea, though.

 

[w00t22]

btw how did you get from
(4)(4^x) + 4^x = 160

to

(5)(4^x) = 160 5?

 

[Ghan]

Treat 4^x as if it's a variable.

We have four of them: (4)(4^x)

and we're adding another:

(4)(4^x) + (4^x)

To get:

(5)(4^x)

The whole package still equals 160:

(5)(4^x) = 160

 

[w00t22]

ah i see makes sense