how to solve an exponential equation with two different bases

We will solve the exponential equation 2^(x+3)=3^(x+2) which has two different bases. This is a must-know algebra problem with exponents and logarithms. For more algebra tutorials, please subscribe to @bprpmathbasics

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"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Feel free to leave your questions in the comment!
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#justalgebra

[militantpacifist]

We need a Just Linear Already and a Just Differential Equations channels with videos.

[htake]

what if we a number with the power x? like 2 to the power 3x

[DownDance]

I came to the blue result (last) before I came to the black one (first), because I instantly used logarithms

[epalegmail]

Why not just transform the bases? 2=3^log3(2) or vice versa, or even using any other basis.

[ianzhou]

I thought of it more intuitively: I took the natural log of both sides, so ln(2^(x+3)) = ln (3^(x+2)). Then, (x+3) ln 2 = (x+2) ln 3. Expanding, we get that x ln 2 + 3 ln 2 = x ln 3 + 2 ln 3. Getting all x's on one side, x ln 2 - x ln 3 = 2 ln 3 - 3 ln 2. Factoring out the x and dividing it by its other factor on the left side, we get the final answer. x = (2 ln 3 - 3 ln 2)/(ln 2 - ln 3).

[abyrinth]

I used logs at the beginning and got to the long answer you got to at the end, then I "simplified" to the single log you originally got as an answer lol

[tavishbryan]

I love doing algebra like this. Once I was first able to do it it was so satisfying coming up with these wild numbers. And now that I've done math for so long it's even more fun because I know so much more about the sets where the numbers live.

[ThePiMan]

Glad I found your just algebra channel sir! Thank you so much!

[ToddKunz]

I absolutely love your videos. Thank you so much.

[lexiette]

I simplified by diving both sides by 2^x+2 to start with leaving me with 2^1 = (3/2)^x+2 which i felt was far easier.

[shadmanhasan]

Don't we just have to take the "ln" on both sides? 🤔
Step 1 -> take "ln" on both sides:
(x+3)ln(2) = (x+2)ln(3)
Step 2 -> cross multiply to isolate "x" from "ln":
(x+3)/(x+2) = ln(3)/ln(2)
Step 3 -> using graphing calculator, find intersection btwn y=(x+3)/(x+2) and y=ln(3)/ln(2) -> which is ~1.585
Intersection: (-0.2905, 1.585)

[OptimusPhillip]

In simplest exact terms: x = log_(2/3)(9/8)

That's the logarithm with base 2/3 of 9/8, in case my syntax is a little confusing.

EDIT: Wow, you handled that a lot more simply than I did. When I saw the x's in the exponential, I automatically applied a natural log to both sides to get it out of there. Then I did a lot of shuffling around to isolate x, and had to apply a bunch of log identities to simplify things down to my final answer.

[asaftk]

easiest way i see is multiply both side by 3, so u get 3x2^(x+3)=3^(x+3), this way both powers r equal and easier to extract as an expression, divide by either 2^(x+3) or 3^(x+3) take log on both sides and solve for x easily.
the expression i got is: [ln(3)/ln(1.5)]-3 value is ofc the same.

[kmsbean]

I got to x = ln(9/8) / (ln2/3) and I thought x = -2, but then I was like, no it isn't, so I stopped

[dumb_tan]

Or you could've just take log 2 both sides, duh!

[harkabirsingh]

X is -2
I don't know how it is just my common sense

[j.s.ospina]

I changed to e^x and ln from the beginning and got log_(3/2)(8/9). Which interestingly enough is the same as log_(2/3)(9/8)

[encounteringjack]

I just did log_2 on both sides and solved for x.

Kind of more simplified in the end.

x = (2log_2(3) - 3)/(1 - log_2(3))