Solve for x: 7^x-1 = 35

Answer 1

To solve for 7^x-1 = 35, we can take the natural log of both sides to use the power rule of logs in order to bring the exponent down since x is in the exponent. The power rule of logs states:

log_ax^p = plog_ax (Notice how the exponent p moved down in the front of log)

So, lets start with taking the natural log of both sides:

ln(7^x-1) = ln(35) This allows us to move the exponent (x-1) down and to the front in the next step

(x-1)ln(7) = ln(35) We can divide by ln(7) on both sides

x - 1 = (ln(35) / ln(7)) Now add 1 on both sides to isolate x

x = (ln(35) / ln(7)) + 1

So, the final answer is (ln(35) / ln(7)) + 1 which is equivalent to 2.827 as a decimal.

Answer 2

start by taking the natural log of both sides

ln 7^x-1 =ln 35 by rules of logs

(x-1) ln7=ln35

x-1 = ln 35/ln7

x= 1 + ln35/ln7