I’m struggling to solve this exponentials problem

[NickKen02]

My apologies if my work is hard to follow, I’m quite stumped on this one, as I haven’t done exp functions work in a while. Thanks in advance for any help.

 

[MarkFL]

I agree with how you began, which is to express everything in powers of 5:

[MATH]\left(\frac{(5^2)^{2x-1}}{(5^3)^{x+3}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(\frac{5^{2(2x-1)}}{5^{3(x+3)}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(\frac{5^{4x-2}}{5^{3x+9}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(5^{x-11}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]5^{2x-22}=5^{\frac{x}{3}}[/MATH]
Now you can equate exponents, and solve for \(x\)...what do you find?

 

[Deleted member 4993]

View attachment 15769
My apologies if my work is hard to follow, I’m quite stumped on this one, as I haven’t done exp functions work in a while. Thanks in advance for any help.

On 3rd line - you lost the "square" on the left-hand-side.

So the equation could be re-written as:

\(\displaystyle \frac{5^{2(2x-1)}}{5^{3(x+3)}} = \ \sqrt[6]{5^x}\)

 

[NickKen02]

I agree with how you began, which is to express everything in powers of 5:

[MATH]\left(\frac{(5^2)^{2x-1}}{(5^3)^{x+3}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(\frac{5^{2(2x-1)}}{5^{3(x+3)}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(\frac{5^{4x-2}}{5^{3x+9}}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]\left(5^{x-11}\right)^2=5^{\frac{x}{3}}[/MATH]
[MATH]5^{2x-22}=5^{\frac{x}{3}}[/MATH]
Now you can equate exponents, and solve for \(x\)...what do you find?

Thanks a lot. I’m kicking myself because it was as simple as remembering to subtract the exponents when put in a situation with an equivalent base and a quotient set up, the answer I got was x=13.2, and after verifying in the original equation it checks out!