Solving exponential equations with different bases

How do we solve exponential equations with different bases? Oh well, make the bases the same first! Sometimes it's easy, sometimes we might have to use a log property that b^logb(x)=x
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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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[@skinnyladd]

for the second equation I did,
(2^4x) x 2 = 3^x
(16^x) x 2 = 3^x
(3^x)/(16^x) = 2
(3/16)^x = 2
Taking log base (3/16) on both sides
x = log base 3/16 (2)
x = -0.414 (which is the same as yours)
great problems, looking forward to more!

[@l.w.paradis2108]

I like how this was explained so clearly and simply. Crisp, without needless complications.

[@quantumnoctemus]

He teaches better in one second than many teachers in 1 hour.

[@neilgerace355]

"Surprising, right?"
Hahahaha

[@inmu5529]

2:00 his humor is crazy

[@real_u23]

so that's when "log that's not base e" is important 😮

[@thomassidoti5496]

I love this guy. It's just funny because if he would have just took the ln of 2 and 3 in the first place he wouldn't have to re-write the answer

[@balduran2003]

It's cool how the answer to the second one still has 1,2,3,4 in it.

[@sanjaybhowmick4905]

I am highly impressed sir

[@tutorchristabel]

Well explained

[@leonardobarrera2816]

That is a good one!!!

[@sebastianjohansen2142]

crazy guy

[@ptrakoo5363]

What is 2 log base2 of 3

[@TheNerdess]

Im supposed to solve using only natural log. that has been my problem finding examples of people solving with that and not log! Stuff like 2^(5x+4)=3^(3x-2) can you just write ln instead of log?

[@carultch]

Since we have the change-of-base rule, it is arbitrary whether you write ln(8)/ln(2) or log(8)/log(2). Both produce the same result. You can solve any problem involving logs, using either natural log or log base ten. Or even some completely different base like log base 2.
To do the problem you provided using natural log:
Given: 2^(5*x+4) = 3^(3*x - 2)
Take the natural log of both sides:
ln(2^(5*x + 4)) = ln(3^(3*x - 2))
Use the log property, ln(a^b) = b*ln(a) to pull the exponents out in front:
(5*x + 4)*ln(2) = (3*x - 2)*ln(3)
Expand, move constants to the right, and variables to the left:
5*x*ln(2) + 4*ln(2) = 3*x*ln(3) - 2*ln(3)
5*x*ln(2) - 3*x*ln(3) = -2*ln(3) - 4*ln(2)
Factor the left:
[5*ln(2) - 3*ln(3)]*x = -2*ln(3) - 4*ln(2)
Isolate x:
x=[-2*ln(3) - 4*ln(2)]/[5*ln(2) - 3*ln(3)]
This can simplify to:
x = -ln(144)/ln(32/27), which evaluates to about -29.25

[@boredafmetoo7467]

Nice Shoes and nice video also

[@mccauleybacalla2228]

wow

[@lookingforahookup]

Isn’t (2^2)^x equal 4^x?

[@ibrahimali3192]

yes

[@graemehumfrey3955]

Beautiful