How do you solve #5^(x-7) = 4#?

Question

Serenity Jones · Accepted Answer

You must convert to logarithmic form and solve for x.

#5^(x - 7) = log4# Simplify using the rule #log_nx^a = alog_nx# #(x - 7)log5 = log4# #xlog5 - 7log5 = log4# #x(log5 - 7log5) = log4# #x = log4/(log5 - log5^7) Using the rule #log_am - log_an = log_a(m / n)# #x = log4/log(5/78125)# Using the rule #loga/logb = log_ba#: #x = log_(1/(15 625))4# Ask your teacher what they require when it comes to the answer. Some teachers like the answer rounded to a certain number of decimals, while some want the exact value answer. Practice exercises: a) #2^(x + 4) = 5# b) #3^(2x - 7) = 2^(x + 1)# c) # 4 xx 2^(x + 2) = 3^(3x - 4)# Good luck!

Christian Antunez · Answer

undefined To solve the equation $5^{x-7} = 4$:

1. Take the natural logarithm (ln) of both sides to eliminate the exponent.
2. Use properties of logarithms to solve for $x$.
3. Solve for $x$ by isolating it.

Starting with the equation $5^{x-7} = 4$:

1. Take the natural logarithm (ln) of both sides:

$\ln(5^{x-7}) = \ln(4)$

2. Use the property of logarithms that allows the exponent to come down as a coefficient:

$(x-7) \cdot \ln(5) = \ln(4)$

3. Solve for $x$ by isolating it:

$x - 7 = \frac{\ln(4)}{\ln(5)}$

$x = \frac{\ln(4)}{\ln(5)} + 7$

Now, you can calculate the value of $x$ using a calculator.