Simultaneous equations (tough one): 8^x 64^y = 128, sqrt[12y - 7] = x

Tuharramah

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Jun 18, 2017
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Hello everyone!
I tried but i couldn't solve it.Even several calculator apps on my phone couldn't. (Photomath .mathlab)
Please explain in an "ok" way because i am not a university student. (Yet :))

8x* 64y =128
Sqrt(12y-7)=x
This is my progress
 

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Hello everyone!
I tried but i couldn't solve it.Even several calculator apps on my phone couldn't. (Photomath .mathlab)
Please explain in an "ok" way because i am not a university student. (Yet :))

8x* 65y =128
Sqrt(12y-7)=x
What have you tried so far. We offer math help here, not solutions.

A hint I will offer is to ask you to find the prime factorization of 128.

Please get back to us with this answer and someone, if necessary, will help you from here.
 
8x* 65y =128
Sqrt(12y-7)=x

This is my progress
View attachment 8160
I cannot read your work. In the future, please crop and rotate your images before posting.

It looks like you wrote 64^y on your paper, instead of 65^y. Does your original post contain typos?

I solved the system that you posted; it involved using properties of logarithms and squaring, to solve each equation for y. Equating the results lead to a quadratic equation. I then switched to decimal approximates, for the coefficients. Checking my final results eliminated an extraneous solution.

The exact solution (found by software) does not simplify nicely, so your original post probably contains a mistake. :cool:
 
Hello everyone!
I tried but i couldn't solve it.Even several calculator apps on my phone couldn't. (Photomath .mathlab)
Please explain in an "ok" way because i am not a university student. (Yet :))

8x* 64y =128
Sqrt(12y-7)=x
This is my progress
18y2 - 60y + 35 = 0

This is a quadratic equation.

You do know how to solve this for y. Correct?
 
I did solve it for y ,Sir.However this is supposed to be solve without a calculator and when I saw decimals as roots I thought I made a mistake somewhere...
 
Did you solve it [using 64y]?
No.

You had posted 65y, so that's what I used.

Can you type out your work, so we can check it?

You can use the carot symbol, to show exponents. :cool:

8^x * 64^y = 128

sqrt(12y - 7) = x
 
Sqrt (12y-7)=x
8^x * 64^y=128

So 2^3x + 2^6y = 2^7

3*sqrt (12y-7) +6y =7

3*(12y-7)=(7-6y)^2

36y-21=49-84y+36y^2

18y^2 - 60y +35 =0
Ans is decimal..
 
I showed one possible solution to his 1st equation.
This was mainly to see if the OP understood.
Did you see their work?


Then used Wolfram ... No results ...Wolfram gave up!
When I tried it, Wolfram showed the solution graphically. If you want to enlarge the graph or see further computations, ya gotta pay.
 
Sqrt (12y-7)=x
8^x * 64^y=128

So 2^3x + 2^6y = 2^7

3*sqrt (12y-7) +6y =7
I wouldn't go this way. Instead:

. . . . .3x + 6y = 7

. . . . .12y - 7 = x^2

Multiply the first equation by -2:

. . . . .-6x - 12y = -14

. . . . .12y - 7 = x^2

Add down to get:

. . . . .-6x - 7 = x^2 - 14

. . . . .0 = x^2 + 6x - 7

. . . . .0 = (x + 7)(x - 1)

...and so forth. :wink:
 
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