Solving an exponential equation with different bases

Here we will solve an exponential equation with different bases. We will solve 2^x=5^(x+2) by using the rules of exponents, logarithm, and the change of base formula. This algebra tutorial is suitable for Algebra 2 students or precalculus students. Subscribe to @bprpmathbasics for more algebra and precalculus tutorials.

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[cyan]

Interesting, I never knew you could do it that way. Here are my steps:
1. Take the natural log on both sides, & got: (x) (In2) = (x+2) (In 5)
2. Distribute: xIn2 = xIn5 + 2In5
3. Subtract xIn5 on both sides: xIn2 - xIn5 = 2In5
4. Factor the x out: x (In2 - In5) = 2In5
5. Divide: x = 2In5 / In2 - In5.
This works everytime for exponential equations with difference bases.

[rafaelivan]

u can actually also just do log both side right away and have:
x log 2= (x+2) log5
x log 2 = x log5 + 2log5
x log 2-x log5= 2log5
x(log2-log5)=2log5
x=2log5/(log2-log5)

[rygerety]

First thought: 2 ^ any power is even and 5^any power is odd. Hmmm

[punteanu]

I enjoy learning math from you early in the morning and late at night, you're cool

[tennesseedarby]

For the last question, answer in the same format:
Step 1: 3^2x = 6^(x - 2)
Step 2: 9^x = (6^x) / 36
Step 3: 36 = (6^x) / (9^x)
Step 4: 36 = (2/3)^x
Step 5: x = logbase2/3(36)
Step 6: x = log(36) / log(2/3)
Step 7: x = 2log(6) / (log(2) - log(3))
Step 8: x = (2log(2) + 2log(3)) / (log(2) - log(3))

[thodoriss]

An important point for students is to note that the reason we can divide by 2^x or 5^x, is that none of these can be equal to zero. Generally, however, when we divide by something that contains an unknown variable, we have to make sure to state it is different than zero and before we give our final answer, we check the case that it is zero.

[geonalugala]

x log 2 = (x + 2) log 5.
x log 2 = x log 5 + 2 log 5
x (log 2 - log 5) = 2 log 5.
x = 2 log 5 /(log 2 - log 5)
= 2 log 5 / log (2/5).

Evaluated:
x = -3.51294159.

[RaspberryMalina]

this has genuinely explained log rules better than my teachers in a few minutes so thank you so much

[adityatrivedi]

Thank you so much! I was just looking for it!

[pirsentheta]

I thank you for proposing these math problems, the solution to the second problem was -2log(6)/(log3-log2)

[abyrinth]

For the question at the end I got 2(log6)/(log2-log3), which can interestingly be rewritten as 2*((log2+log3)/(log2-log3))

[devinmurphy]

What’s interesting about this is that when you get to (2/5)^x=25. Intuitively, whatever number you choose for x will affect the fraction of 2/5. But the larger the number x is, the faster the 5 will grow in comparison to the 2. And you will never be able to find a positive value for x where 2^x/5^x=25 because the denominator will always be bigger. Then if we use exponent rules, we can realize that our x value will have to be negative. A negative x will swap the numerator and denominator and allow for the fraction to be equivalent, for some negative x value, to 25

[peamutbubber]

I thought of 2^x as e^xln2 and 5^x+2 as e^(x+2)ln5 and arrived to the same answer! Thanks to your previous teaching of course

[maxysa]

Linguagem matemática é universal mesmo, entendi tudo sem entender uma palvra, ótima explanação. Grata.

[jessetrevena]

The answer I got for the last question is -2ln(6)/ln(9)-ln(6), the exact answer is -8.838045165. In my opinion taking the natural log
of both sides for questions like this is very quick and easy to manipulate the equation to find x.

[AlideEricsonMwandila]

This is so amazingly that everyone shouldn't complain about solving exponential logarithm with diff bases, but to be thankful...Thank you sir for this assistance once again.

[matheuslopes]

The coolest part Is always the cute Little Pokéball in his hand. It's such a nice touch!

[d.n.ctalop]

I represented 6 as 3^log3(6), and compared the powers. X ≈ -8.83

[ESOMNOFUONLINEMATH]

I have been your subscriber for over 3 years now.
You inspired me to start my own channel

[omaraborgela]

We can simply make
Ln to the both sides so the exponents will get down
x.ln2=(x+2).ln5