2^6x-1 = 9
and...
can you solve exponential functions without using log if the base isn't the same..?
log(2^(6x-1))=log9, (assuming your exponent is 6x-1)
(6x-1)log2=log9
6x-1=log9/log2
etc
Unless you use some kind of interpolation, using logs along with a calculator will give you the most accurate answer.
(I got x=.6949875)
you got 8 hours of sleep on Monday night and 7 hours of sleep on tuesday night. On wednesday, you got 4 hours of sleep because you were up late working on a english paper. On thursday and friday you got 9 hours of sleep each night to make up for wednseday. Use a verbal model to find how many hours of sleep you got during the week?
To solve the equation 2^(6x-1) = 9, we can use logarithms. However, if the base is not the same, we cannot solve it directly without using logarithms. Let me explain how we can use logarithms to solve this equation.
1. Take the logarithm of both sides: log(2^(6x-1)) = log(9). Since the base is not explicitly mentioned, we can assume it as 10 for simplicity.
2. By the logarithmic property log(base^exponent) = exponent * log(base), we can rewrite the left side as (6x-1) * log(2).
3. Now the equation becomes (6x-1) * log(2) = log(9).
4. Divide both sides by log(2) to isolate 6x-1: (6x-1) = log(9) / log(2).
5. Simplify the right side by finding the logarithms: (6x-1) ≈ 3.169925 / 0.30103. (Approximate values are used here for illustration purposes).
6. Solve for x by adding 1 to both sides: 6x ≈ 3.169925 / 0.30103 + 1.
7. Finally, divide both sides by 6 to obtain the value of x: x ≈ (3.169925 / 0.30103 + 1) / 6.
So, x is approximately equal to the calculated value.
In summary, without using logarithms, solving equations where the bases are not the same can be quite complicated. Logarithms are valuable tools in such cases as they allow us to transform exponential equations into simpler linear equations that can be solved more easily.