using logarithms to solve exponential equations.
5^1+x = 2^1-x
I need exact numbers.
I did one on my own already. 5^x-1 = 9
5^x-1 = 9
log(5^x-1) = log9
(log5)(x-1) = log9
x-1 = (log9/log5)
x= (log9/log5)-1
x = 2.3652
I really need help
log(5^(1+x))= log(2^(1-x))
(1+x)(log 5)= (1-x)(log 2)
(x+1)/(x-1)=.4306765581
x+1= .4306765581x-.4306765581
At this point it's a simple algebraic solution to solve for x. :)
The answer in the textbook says -0.398. I don't think the answer above gives me -0.398.
(1+x)(log 5)= (1-x)(log 2)
(1+x)/(1-x) = log2/log5 = .4306765581
x = -0.39794
To solve the equation 5^(1+x) = 2^(1-x) using logarithms, follow these steps:
Step 1: Take the logarithm of both sides of the equation. Since the bases of the exponents are different, you can choose any logarithm base. Let's use the common logarithm (base 10).
Take the logarithm of both sides:
log(5^(1+x)) = log(2^(1-x))
Step 2: Use the logarithmic property that states log(a^b) = b * log(a) to simplify the equation.
Apply the logarithmic property to the left side:
(1+x) * log(5) = (1-x) * log(2)
Step 3: Distribute the logarithm to both terms on the right side of the equation.
Expand the equation:
log(5) + x * log(5) = log(2) - x * log(2)
Step 4: Move the x terms to one side of the equation and the constants to the other side to isolate x.
Rearrange the equation:
x * log(5) + x * log(2) = log(2) - log(5)
Factor out x on the left side:
x * (log(5) + log(2)) = log(2) - log(5)
Step 5: Simplify the logarithmic expressions on both sides of the equation.
Combine the logarithmic expressions:
x * log(10) = log(2/5)
Since log(10) = 1, the equation simplifies to:
x = log(2/5)
Step 6: Calculate the value of x using a calculator.
Using a calculator, evaluate the logarithm of 2/5:
x = log(2/5) ≈ -0.3979
Therefore, the exact solution to the equation 5^(1+x) = 2^(1-x) is x ≈ -0.3979.