solve eqn w/ variable in exponent using logarithms: 10*2^x - 8*5^(x-2) = 0

[Ghans]

thank you for your help in advance

it is problem number 10 below. if there is any way to show the steps I really appreciate it. the answer is supposed to be (log 31.25/log 2.5) which approxiately equals 3.75

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$\displaystyle \textbf{10. }\quad 10\, \cdot\, 2^x\, -\, 8\, \cdot\, 5^{x-2}\, =\, 0$

 

[Deleted member 4993]

thank you for your help in advance

it is problem number 10 below. if there is any way to show the steps I really appreciate it. the answer is supposed to be (log 31.25/log 2.5) which approxiately equals 3.75

---

$\displaystyle \textbf{10. }\quad 10\, \cdot\, 2^x\, -\, 8\, \cdot\, 5^{x-2}\, =\, 0$

solve:

10 * 2^x - 8 * 5^(x-2) = 0

2^(-2) * 2^x - 5^(-1) * 5^(x-2) = 0

Continue.....

 

[Ghans]

thank you! I solved for the answer

however, I am helping someone who has more trouble with math. Is there an alternative method to solve the problem which might require less insight?

 

[lookagain]

the answer is supposed to be (log 31.25/log 2.5) which approxiately equals 3.75

$\displaystyle \dfrac{log(31.25)}{log(2.5)} \ = \ 3.756...$


It approximately equals 3.76.

 

[JeffM]

thank you for your help in advance

it is problem number 10 below. if there is any way to show the steps I really appreciate it. the answer is supposed to be (log 31.25/log 2.5) which approxiately equals 3.75

---

$\displaystyle \textbf{10. }\quad 10\, \cdot\, 2^x\, -\, 8\, \cdot\, 5^{x-2}\, =\, 0$

An alternative approach

$\displaystyle 10 * 2^x - 8 * 5^{(x - 2)} = 0 \implies 10 * 2^x = 2^3 * 5^{(x - 2)} \implies$

$\displaystyle log_{10}(10 * 2^x) = log_{10}(2^3 * 5^{(x - 2)}) \implies WHAT?$

Hint: After simplifying logarithms, gather terms involving x on LHS and terms not involving x on RHS.

 

[Ghans]

thanks everyone!

 

[tkhunny]

thank you! I solved for the answer

however, I am helping someone who has more trouble with math. Is there an alternative method to solve the problem which might require less insight?

Kind of an offensive question.

Yes, there is. Factor everything. Prime factorization.

10 = 2*5

Collect all the like factors and look at it again.

 

[Ghans]

Kind of an offensive question.

believe me, no offense intended

it has been approx 40 years since taking the class and I'm quite rusty. made a stupid mistake when figuring out the answer yesterday, was stumped, and thus posted here. this group straightened me out quickly

thanks again for all the help

 

[Ghans]

[quote


it approximately equals 3.76.[/quote]


+1

 

[Steven G]

thank you for your help in advance

it is problem number 10 below. if there is any way to show the steps I really appreciate it. the answer is supposed to be (log 31.25/log 2.5) which approxiately equals 3.75



View attachment 9508

I would not solve this 1st one by using logs. 2^(2x+5) = (1/2) = 2^-1. So now solve 2x+5 = -1

For the 2nd problem I would also try to solve without logs:

10*2^x = 8*5^x-2
5*2*2^x = 2³*5^x-2
5*2^x+1 = 2³*5^x-2
Then I would equate 1 and x-2 as well as x+1 and 3

1 = x-2 so x=3 and x+1 = 3 so x=2. Hmm, we got different answers for x so we have to use logs

 

[tkhunny]

believe me, no offense intended

it has been approx 40 years since taking the class and I'm quite rusty. made a stupid mistake when figuring out the answer yesterday, was stumped, and thus posted here. this group straightened me out quickly

thanks again for all the help

None taken.