Advanced Differential Equation Problem: 4^x * 5^4x+3 = 10^2x+3

[Onigma]

Hi,

I need some help solving this equation using logarithms: 4^x * 5^4x+3 = 10^2x+3

I know the answer is x = log25 8, I just have no idea how to work it out,
and the problem uses the change of base rule: logba= (logca) / (logcb)

Thanks in advance : )

 

[mmm4444bot]

Advanced Differential Equation …

I need some help solving this equation using logarithms: 4^x * 5^(4x+3) = 10^(2x+3)

I know the answer is x = log₂₅(8)

I just have no idea how to work it out, and the problem uses the change of base rule:

log_b(a) = log_c(a) / log_c(b)

This is not a differential equation. It's a high-school level algebra exercise dealing with properties of exponents and logarithms. What class are you taking?

Also, note the added grouping symbols (in red) around the binomial exponents; if you don't type those grouping symbols, then it's a different equation with a different answer. (I was able to figure out what you intended only because you posted a solution.)

What have you already tried? Where did you get stuck?

If you can't begin, try using properties of exponents, to rewrite everything as base 2 and base 5 powers. You can then separate those powers (i.e., get the base 2 powers to one side of the equation and the base 5 powers to the other side). This results in ratios. There's another property of exponents that simplifies a ratio of powers with the same base.

4^x = 2^?

10^(2x+3) = (2*5)^(2x+3) = 2^? * 5^?

Please reply with your work, so we can see how far you get and check it.

If you can't remember the properties of exponents, google them (or check your textbook's index).

Please also read the forum guidelines. Thank you! :cool:

 

[stapel]

I need some help solving this equation using logarithms: 4^x * 5^4x+3 = 10^2x+3

I know the answer is x = log25 8, I just have no idea how to work it out,
and the problem uses the change of base rule: log_ba = (log_ca) / (log_cb)

I will assume that the exercise is actually as follows:

. . . . .$\displaystyle 4^x\, 5^{4x+3}\, =\, 10^{2x + 3}$

Try using what you learned back in pre-algebra about powers. Is there any way to relate the bases, 4 (which is the square of 2), 5, and 10? (Hint: Multiplication fact.)

How can the left-hand side be re-stated as the product of a power on 2 and a power on 5?

How can the right-hand side be restated as a power on 10?

How then can it be restated as the product of a power on 2 and a power on 5?

Simplifying the resulting equation (by dividing off equal factors), with what are you left?

When you solve this for "x=", where does this lead?

If you get stuck, please reply showing all of your steps so far. Thank you!

 

[Onigma]

My apologies for not reading the forum guidelines and posting in the wrong section.

Well thank you much for y'all's help. With what you told me I was able to complete the question. Thanks a bunch!