Two Exponential Equations – Thoughts Your Selections

Due to Don for alerting me of a serious typo–the complete downside 2 was incorrectly written at first.

Downside 1

Remedy for all actual values of x such that

9^x + 12^x = 16^x

Downside 2

Remedy for all actual values of x such that

(8^x – 2^x)/(6^x – 3^x) = 2

As typical, watch the video for an answer.

Two Exponential Equations

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“All shall be properly in case you use your thoughts in your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of recreation concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts due to neighborhood assist! Assist out and get early entry to posts with a pledge on Patreon.

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Reply To Exponential Equation 8 to x minus 2 to x

(Just about all posts are transcribed shortly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).

First divide every time period by 9^x

9^x + 12^x = 16^x
9^x/9^x + 12^x/9^x = 16^x/9^x
1 + (12/9)^x = (16/9)^x
1 + (4/3)^x = ((4/3)²)^x
1 + (4/3)^x = (4/3)^2x

Then let u = (4/3)^x so the equation reduces to a quadratic equation.

1 + u = u²
0 = u² – u – 1

Utilizing the quadratic components we get two roots

u = (1 ± √(1 – 4(1)(-1)))/2

u = (1 + √5))/2
u = (1 – √5))/2

Since u = (3/2)^x, that is larger than 0 for actual values of x. So we reject the second root which is unfavourable. Thus:

u = (1 + √5))/2

So u is the same as the golden ratio! Now we have now one step left: to resolve for x. We are able to do this with the assistance of the pure logarithm.

u = (4/3)^x
ln u = ln[(4/3)^x]
ln u = x ln(4/3)
x = ln u/(ln (4/3))
x = [ln((1 + √5)/2)]/(ln (4/3))
x ≈ 1.673

The equation is difficult at first. However with just a few intelligent methods it’s attainable to discover a golden reply!

Downside 2

This equation is far more durable and we sadly can not use the identical trick as in downside 1. So let’s simplify it fastidiously and examine its habits.

Clearly x ≠ 0 as a result of the fraction could be undefined:

(8⁰ – 2⁰)/(6⁰ – 3⁰)
= (1 – 1)/(1 – 1)
= 0/0

Additionally x = 1 is an answer

(8¹ – 2¹)/(6¹ – 3¹)
= (8 – 2)/(6 – 3)
= 4/2
= 2

Are there another actual options? The reply isn’t any!

If x ≠ 0, then we have now:

(8^x – 2^x)/(6^x – 3^x)
= (2^3x – 2^x)/(2^x3^x – 3^x)
= 2^x(2^2x – 1)/[3^x(2^x – 1)]
= 2^x(2^x – 1)(2^x + 1)/[3^x(2^x – 1)]

Since x will not be 0, we are able to cancel 2^{x – 1} within the numerator and denominator.

2^x(2^x + 1)/3^x
= (2^2x + 2^x)/3^x
= (4^x + 2^x)/3^x
= (4/3)^x + (2/3)^x

Outline a operate for all actual values of x:

f(x) = (4/3)^x + (2/3)^x – 2

Each resolution of the unique fraction shall be a root of this equation. What are the roots of this equation? Calculate the second by-product:

f”(x) = (4/3)^x(ln (4/3))² + (2/3)^x(ln (2/3))²

Each time period is strictly larger than 0, so the complete 2nd by-product is strictly larger than 0. This implies the unique operate is convex and has at most 2 distinct roots.

Clearly x = 1 is a root. However we even have x = 0 is a root:

(4/3)⁰ + (2/3)^x – 2
= 1 + 1 – 2
= 0

Each root of the unique fraction equation could be a root of this operate. However this operate has two roots x = 1 and x = 0. Since x ≠ 0 within the authentic fraction, the one actual root is x = 1.

References

Just like downside 1 on Socratic
https://socratic.org/questions/how-do-you-solve-6-x-4-x-9-x

Downside 2 @MathBooster video integer options
https://www.youtube.com/watch?v=X3OiHW98QTE

Downside 2 Math StackExchange
https://math.stackexchange.com/questions/4699669/frac8x-2x6x-3x-2-solve-for-x

Downside 2 strictly convex
https://math.stackexchange.com/questions/2112108/prove-a-convex-and-concave-function-can-have-at-most-2-solutions

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