# the base of exponentialsolve 14^(14x+1)=5^x, but the bases are not equal. the eq. has solutions?

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

14^(14x+1) = 5^x

To solve, we will apply the logarithm to both sides.

==> log 14^(14x+1) = log 5^x

Now we will use the logarithm properties to solve.

We know that log a^b  = b^log a

==> (14x+1)*log 14 = x *log 5

Now we will divide by log14*x

==> (14x+1)/x = log5/log14

==> 14 + 1/x = log5/log14

==> 1/x = log5/log14 - 14

= -13.39

==> x = 1/-13.39 = -0.0746

==> Then, the solution for the equation is x = -0.0746

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since the bases are primes and they are not matching, we'll use logarithms to solve exponential equations.

We'll take logarthims both sides. The base of logarithm is the base of exponential that has the variable to superscript.

log14  [14^(14x+1)] = log14 (5^x)

We'll apply the power rule for logarithms:

(14x+1) log14 14 = x log14 5

We'll recall that log14 14 = 1

We'll re-write the equation:

14x+1 = x log14 5

We'll subtract x log14 5  both sides to isolate x terms to the left side.:

14x - x log14 5 = -1

We'll factorize by x:

x(14 - log14 5) = -1

We'll re-write log14 5 = lg5/lg14

x(14 - lg5/lg14) = -1

We'll divide by 14 - lg5/lg14 = 13.3901

x = -1/13.3901

Rounded to four decimal places:

x = -0.0746

So, the equation has the solution x = -0.0746.