SOLUTION: How would you solve the equation .25^x-0.5=2 ?
Algebra.Com
Question 294959: How would you solve the equation .25^x-0.5=2 ?
Answer by helpnalgebra(91) (Show Source): You can put this solution on YOUR website!
25^(x)-0.5=2
25^(x)=0.5+2
25^(x)=2.5
ln(25^(x))=ln(2.5)
xln(25)=ln(2.5)
xln(25)=(0.92)
(xln(25))/(ln(25))=(0.92)/(ln(25))
x=(0.92)/(ln(25))_xAPPR0.28 answer
RELATED QUESTIONS
How would you solve problems like:
x^2-25=0 x=....
x^2-49=0 x=....
Thank... (answered by bsmath)
I forgot to give an e-mail address for the question: How would you factor this:... (answered by Earlsdon,Photonjohn)
I forgot to give an e-mail address for the question: How would you factor this:... (answered by Earlsdon)
How do you solve the equation x^2-6x-25=0 and write your answer in the form ab+i? (answered by stanbon)
How would you solve 5^(x+1) =... (answered by robertb)
How would you solve the equation
4e^x+1=5 (answered by addingup,ikleyn,MathTherapy)
can you square a negative number?
im trying to solve the equation... (answered by RAY100)
How would you solve this equation?... (answered by jim_thompson5910)
How do you factor out the equation a^2 - 25 = 0? Help would be appreciated. Thank... (answered by Alan3354)