# how wod i solve the following two eqautions? either exactly or to 3 significant

a)2^x = 5

b)3^x+1 = 7

i know a) is 2.32 and b) is 0.771 but i don't know how to show it out step by step in how i worked it out .

can sum1 plz help me

You have to use logs, and learn how to do that on your calculator.

3^(x+1)=7

x+1= log3 7

## To solve the equation 3^(x+1) = 7, we can start by taking the logarithm of both sides. In this case, we will use the natural logarithm (ln) because it is commonly used in solving exponential equations.

So, we have:

ln(3^(x+1)) = ln(7)

Next, we can use the logarithm property, which states that ln(a^b) = b * ln(a). Applying this property to the left side of the equation, we have:

(x+1) * ln(3) = ln(7)

Now, let's solve for x by isolating it. Divide both sides of the equation by ln(3):

(x+1) = ln(7) / ln(3)

Finally, subtract 1 from both sides:

x = ln(7) / ln(3) - 1

Now, we can evaluate the expression on the right side using a calculator. The calculated value for x will be approximately 0.770965.

For equation a) 2^x = 5, we can follow a similar process:

Take the logarithm of both sides using the base 10 logarithm (log):

log(2^x) = log(5)

Again, use the logarithm property to simplify the equation:

x * log(2) = log(5)

Isolate x by dividing both sides by log(2):

x = log(5) / log(2)

Now, evaluate the expression on the right side using a calculator. The calculated value of x will be approximately 2.32193.

In summary, to solve exponential equations like these, you can use logarithms to convert the equation into a form that allows you to solve for the variable.