- #1
chjopl
- 21
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2^x=7*5^x
I know you have to take ln of both sides but i am having trouble doing that for the right side.
I know you have to take ln of both sides but i am having trouble doing that for the right side.
chjopl said:2^x=7*5^x
I know you have to take ln of both sides but i am having trouble doing that for the right side.
The first step in solving this equation is to rewrite it as 2^x = 7^(1-x).
The most common mistake is forgetting to distribute the exponent of x to both 7 and 5, resulting in an incorrect solution.
You can simplify this equation by dividing both sides by 5^x, resulting in 2^(x-x)=7, which simplifies to 2^0=7, or 1=7. Since this is not a true statement, there is no solution.
Yes, there are restrictions when the bases of the exponents are equal. In this case, the equation would simplify to 2^x=7*2^x, which has no solution. This is because the bases must be different in order for the equation to be solvable.
Yes, you can use logarithms to solve this equation. Taking the logarithm of both sides will result in x*log2 = log7 + x*log5. You can then rearrange the equation to solve for x.