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Solve for x 4^x-2=256

User Jean Vitor
by
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2 Answers

5 votes

4^(x - 2) = 256 \\ln(4^(x - 2)) = ln(256) \\(x - 2)ln(4) = ln(256) \\((x - 2)ln(4))/(ln(4)) = (ln(256))/(ln(4)) \\x - 2 = (ln(2^(8)))/(ln(2^(2))) \\x - 2 = (8ln(2))/(2ln(2)) \\x - 2 = 4 \\x = 6
User Radek Lopatecki
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7.7k points
7 votes
We can do this problem using two ways....
Let me show you both methods.

Method 1: "Without using logs"


4^(x-2)=256\\\\ 4^(x-2)=4^4\ \ \ \ \ |\because 4^4 \ $is the same as 256

Now, since both sides of the equation have same base and they equal each other, then their exponent must be equal to each other. So, we equal their exponents.

After equaling the exponents, we get,

x - 2 = 4 \\\\ x=4+2 \\\\ \boxed{x = 6} \Leftarrow $ This is your answer!!

Now,
Method 2: "Using Logs"


4^(x-2)=256 \\\\ ~~~~~~~\Downarrow ~~~~~~ is\ the\ same\ thing\ as\ saying \\\\$log\ _4\ 256 =x-2 \\\\ Now,\ using\ the\ log\ property\ log_AB= (log_cA)/(log_cB), \ we \ get, \\\\ (log_(10)256)/(log_(10)4)=x-2 \\\\ 4=x-2 \\\\ 4+2=x\\\\ 6=x \\\\ \boxed{x=6} \Leftarrow $This is your answer



So, either way we get x= 6
User Edilio
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8.4k points