## Problem 875

Write the equation in its equivalent logarithmic form.

73 = 343

Solution:-

The equivalent logarithmic form

3 =

## Problem 874

Write the following equation in its equivalent exponential form.

= y

Solution:-

The exponential form is 4y = 64

## Problem 873

Write the equation in its equivalent exponential form.

2 =

Solution:-

The exponential form is b2 = 49.

## Problem 872

Write the equation in its equivalent exponential form,

2 =

Solution:-

The  equivalent exponential form

32 = 9

## Problem 871

Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.

ex+11 = .

Solution:-

An exponential equation is an equation containing a various in an exponent. Some exponential equations can be solved by expressing each side of the equation as a power of the same base.

All exponential functions are one-to-one. That is, two different ordered pairs have the same second component. Thus, if is a positive number other than 1 and bM = bN, THEN M = N.

To solve the exponential equation ex+11 = , first express both sides of the equation as a power of the same base.

e x+11  =

e x+11  = e-1

Then, set the exponents equal to each other and solve for x.

x + 11 = -1

x = -12

Therefore, the solution set is {-12}.

## Problem 870

Solve the exponential equation by expressing each side as a power of the same base and then equating exponents.

2x = 8192

Solution:-

To solve the exponential equation, start by expressing each side as the power of the same base. Since the base on the side is 2, use the guess and check method to determine the power of that equals the right side.

Now express both sides of the equation as the power of the same base.

2x = 8192

2x = 213

x = 13

Therefore, the solution set is {13}.

## Problem 733

Evaluate 16log1629

Solution:-

Use the following property of  logarithms to evaluate the given if a > 0 and a ≠ 1.

alogax = x (x >0)

Determine the value for a in the given expression.

a = 16

Now determine the value for x in the given expression.

x = 29

alogax = x

16log1629 = 29

Therefore, the value of 16log1629  is 29.

## Problem 732

Use the properties of logarithms to evaluate the expression.

log 1015

Solution:-

To simplify this expression, use the logarithmic property log bbx=x. Check to see if the given expression matches the form log b bx.

The base of the given logarithmic expression is 10.

To simplify the logarithm, base 10, of  10 to a power, use the above logarithmic property. In the expression log 1015, b = 10 and x = 15.

Log 1015  = 15

## Problem 731

Write log aa = 15x in exponential form and find x to evaluate log aa for any a > 0, a ≠ 1.

Solution:-

Recall y = logb x  is called the logarithmic form of the equation and x = by is called the exponential form of the equation. The number b is called the base in both in both y = log b x and x = by , and y is the logarithm in y = log b x and the exponent in x = by. Thus, a logarithm is an exponent.

To write log aa = 15x in exponential form, the variable a is treated as the base.

The expression 15x is the exponent.

Therefore, the exponential form of the equation log aa = 15x is a15x = a.

Now, to find the value of x, set the exponents on both sides of the equation, a15x = 0, equal and solve.

15x = 1

x =

Recall the log aa = 15x and x  =. Substitute the value of x in the equation log aa = 15x and evaluate log aa, where a > 0 , a ≠ 1.

log aa = 15()

log aa = 1

Thus, the exponential form of the equation log aa = 15x is a15x = a and x = for any a > 0, a ≠ 1.

## Problem 730

Write the inverse of y = 14x in logarithmic form.

Solution:-

To find the inverse of y = 14x, first rewrite the equation by interchanging x  and y.

x = 14y

Solve the new equation for y . Here, y is the power to which 14 raised to get the number x.

For x > 0, b > 0, and b ≠ 1, the logarithmic function to the base b is y = log bx, which is defined by x = by.

Now write the inverse in logarithmic form.

y = log 14x

Therefore, the inverse of y = 14x in logarithmic form is y = log 14x.